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NCERT Solutions For Class 10 Maths Chapter 14-Statistics

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics are provided here, which can be downloaded for free, in PDF format. The solutions are prepared by our subject experts who have mastery in Maths. All the solved questions of Statistics are with respect to the latest updates on the CBSE syllabus and guidelines, to help students solve each exercise question and effectively prepare for the CBSE exam.

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Using these solutions as a reference tool will be helpful for the students to score good marks. Students can also get the exercise-wise Solutions for Class 10 Maths in all chapters and practise solving the problems.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 15 Probability

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

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Access Answers to Maths NCERT Class 10 Chapter 14 – Statistics

Exercise 14.1 Page: 270

1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?

To find the mean value, we will use the direct method because the numerical value of f i and x i are small.

Find the midpoint of the given interval using the formula.

Midpoint (x i ) = (upper limit + lower limit)/2

The formula to find the mean is:

Mean = x̄ = ∑f i x i /∑f i

Therefore, the mean number of plants per house is 8.1.

2. Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method .

In this case, the value of mid-point (x i ) is very large, so let us assume the mean value, a = 550.

Class interval (h) = 20

So, u i = (x i – a)/h

u i = (x i – 550)/20

Substitute and find the values as follows:

So, the formula to find out the mean is:

Mean = x̄ = a + h(∑f i u i /∑f i ) = 550 + [20 × (-12/50)] = 550 – 4.8 = 545.20

Thus, mean daily wage of the workers = Rs. 545.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

To find out the missing frequency, use the mean formula.

Given, mean x̄ = 18

The mean formula is

Mean = x̄ = ∑f i x i /∑f i = (752 + 20f)/ (44 + f)

Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752 + 20f)/ (44 + f)

⇒ 18(44 + f) = (752 + 20f)

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

From the given data, let us assume the mean as a = 75.5

x i = (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the u i and f i u i as follows:

Mean = x̄ = a + h(∑f i u i /∑f i )

= 75.5 + 3 × (4/30)

= 75.5 + (4/10)

= 75.5 + 0.4

Therefore, the mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1.

Here, assumed mean (a) = 57

Here, the step deviation is used because the frequency values are big.

The formula to find out the Mean is:

= 57 + 3(25/400)

= 57 + 0.1875

Therefore, the mean number of mangoes kept in a packing box is 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Let us assume the mean (a) = 225

Class size (h) = 50

= 225 + 50(-7/25)

= 225 – 14

Therefore, the mean daily expenditure on food is 211.

7. To find out the concentration of SO 2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO 2 in the air.

To find out the mean, first find the midpoint of the given frequencies as follows:

The formula to find out the mean is

= 0.099 ppm

Therefore, the mean concentration of SO 2 in the air is 0.099 ppm.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

The mean formula is,

= 12.48 days

Therefore, the mean number of days a student was absent = 12.48.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean

literacy rate.

In this case, the value of mid-point (x i ) is very large, so let us assume the mean value, a = 70.

Class interval (h) = 10

u i = (x i – 70)/10

So, Mean = x̄ = a + (∑f i u i /∑f i ) × h

= 70 + (-2/35) × 10

Therefore, the mean literacy part = 69.43%

Exercise 14.2 Page: 275

1. The following table shows the ages of the patients admitted to a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two

measures of central tendency.

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 23, so the modal class = 35 – 45,

Lower limit of modal class = l = 35,

class width (h) = 10,

f 1 = 21 and f 2 = 14

The formula to find the mode is

Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

= 35 + (20/11)

= 36.8 years

So the mode of the given data = 36.8 years

Calculation of Mean:

First find the midpoint using the formula, x i = (upper limit +lower limit)/2

= 35.375 years

Therefore, the mean of the given data = 35.375 years

2. The following data gives the information on the observed lifetimes (in hours) of 225

electrical components:

Determine the modal lifetimes of the components.

From the given data the modal class is 60–80.

Lower limit of modal class = l = 60,

The frequencies are:

f m = 61, f 1 = 52, f 2 = 38 and h = 20

Mode = l+ [(f m – f 1 )/(2f m – f 1 – f 2 )] × h

Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20

Mode = 60 + (45/8) = 60 + 5.625

Therefore, modal lifetime of the components = 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200

families of a village. Find the modal monthly expenditure of the families. Also, find the

mean monthly expenditure:

Given data:

Modal class = 1500-2000,

Frequencies:

f m = 40 f 1 = 24, f 2 = 33 and

Mode formula:

Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500

Mode = 1500 + (8000/23) = 1500 + 347.83

Therefore, modal monthly expenditure of the families = Rupees 1847.83

Calculation for mean:

First find the midpoint using the formula, x i =(upper limit +lower limit)/2

Let us assume a mean, (a) be 2750.

The formula to calculate the mean,

Mean = x̄ = a +(∑f i u i /∑f i ) × h

Substitute the values in the given formula

= 2750 + (-35/200) × 500

= 2750 – 87.50

So, the mean monthly expenditure of the families = Rs. 2662.50

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures

Modal class = 30 – 35,

Class width (h) = 5,

f m = 10, f 1 = 9 and f 2 = 3

Mode Formula:

Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5

= 30 + (5/8)

= 30 + 0.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, x i =(upper limit +lower limit)/2

= 1022.5/35

= 29.2 (approx)

Therefore, mean = 29.2

5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.

Find the mode of the data.

Modal class = 4000 – 5000,

class width (h) = 1000,

f m = 18, f 1 = 4 and f 2 = 9

Substitute the values

Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000

= 4000 + (14000/23)

= 4000 + 608.695

= 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

Given Data:

Modal class = 40 – 50, l = 40,

Class width (h) = 10, f m = 20, f 1 = 12 and f 2 = 11

Mode = l + [(f m – f 1 )/(2f m – f 1 – f 2 )] × h

Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10

= 40 + (80/17)

Thus, the mode of the given data is 44.7 cars.

Exercise 14.3 Page: 287

1. The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

Find the cumulative frequency of the given data as follows:

From the table, it is observed that, N = 68 and hence N/2=34

Hence, the median class is 125-145 with cumulative frequency = 42

Where, l = 125, N = 68, cf = 22, f = 20, h = 20

Median is calculated as follows:

= 125 + [(34 − 22)/20] × 20

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

f m or f 1 = 20, f 0 = 13, f 2 = 14 & h = 20

Mode = l+ [(f 1 – f 0 )/(2f 1 – f 0 – f 2 )] × h

Mode = 125 + [(20 – 13)/ (40 – 13 – 14)] × 20

= 125 + (140/13)

= 125 + 10.77

Therefore, mode = 135.77

Calculate the Mean:

x̄ = a + h (∑f i u i /∑f i ) = 135 + 20 (7/68)

Mean = 137.05

In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5, find the value of x & y.

Given data, n = 60

Median of the given data = 28.5

Where, N/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class, l = 20,

cf = 5 + x,

f = 20 & h = 10

28.5 = 20 + [(30 − 5 − x)/20] × 10

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8.

Now, from cumulative frequency, we can identify the value of x + y as follows:

60 = 45 + x + y

Now, substitute the value of x, to find y

60 = 45 + 8 + y

y = 60 – 53

Therefore, the value of x = 8 and y = 7.

3. The life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.

Given data: N = 100 and N/2 = 50

Median class = 35-40

Then, l = 35, cf = 45, f = 33 & h = 5

Median = 35 + [(50 – 45)/33] × 5

= 35 + (25/33)

Therefore, the median age = 35.76 years.

4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:

Find the median length of the leaves.

(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180.5.)

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

So, the data obtained are:

N = 40 and N/2 = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

Median = 144.5 + [(20 – 17)/ 12] × 9

= 144.5 + (9/4)

= 146.75 mm

Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a lifetime of 400 neon lamps.

Find the median lifetime of a lamp.

N = 400 & N/2 = 200

Median class = 3000 – 3500

Therefore, l = 3000, cf = 130,

f = 86 & h = 500

Median = 3000 + [(200 – 130)/86] × 500

= 3000 + (35000/86)

= 3000 + 406.98

Therefore, the median lifetime of the lamps = 3406.98 hours

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

To calculate median:

N = 100 & N/2 = 50

Median class = 7-10

Therefore, l = 7, cf = 36, f = 40 & h = 3

Median = 7 + [(50 – 36)/40] × 3

Median = 7 + (42/40)

Median = 8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f 1 = 40, f 0 = 30, f 2 = 16 & h = 3

Mode = 7 + [(40 – 30)/(2 × 40 – 30 – 16)] × 3

= 7 + (30/34)

Therefore mode = 7.88

Mean = 832/100 = 8.32

Therefore, mean = 8.32

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Given: N = 30 and N/2= 15

Median class = 55-60

l = 55, C f = 13, f = 6 & h = 5

Median = 55 + [(15 – 13)/6] × 5

= 55 + (10/6)

= 55 + 1.666

Therefore, the median weight of the students = 56.67

Exercise 14.4 Page: 293

1. The following distribution gives the daily income of 50 workers in a factory.

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Convert the given distribution table to a less than type cumulative frequency distribution, and we get

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them to get a smooth curve. The curve obtained is known as less than type ogive.

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the median by making a table.

Here, N = 35 and N/2 = 35/2 = 17.5

Median class = 46 – 48

Here, l = 46, h = 2, cf = 14, f = 14

The mode formula is given as:

= 46 + [(17.5 – 14)/ 14] × 2

= 46 + 0.5 = 46.5

Thus, median is verified.

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution and draw its ogive.

Converting the given distribution to a more than type distribution, we get

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on the graph paper. The graph obtained is known as more than type ogive curve.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Class 10 Maths Chapter 14, Statistics, is one of the most important chapters present in the textbook. The weightage of this chapter in the CBSE exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4 (it could vary as per question).

Topics covered in Chapter 14, Statistics are as follows:

• Mean of Grouped Data
• Mode of Grouped Data
• Median of Grouped Data
• Graphical Representation of Cumulative Frequency Distribution

List of Exercises in Class 10 Maths Chapter 14 : Exercise 14.1 Solutions 9 Questions ( 9 long) Exercise 14.2 Solutions 6 Questions ( 6 long) Exercise 14.3 Solutions 7 Questions ( 7 long) Exercise 14.4 Solutions 3 Questions ( 3 long)

NCERT solutions for Class 10 Maths Chapter 14 – Statistics are made available for students aiming to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions , such that, students find it easy to understand the fundamentals quickly.

The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.

You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form, in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the CBSE exam.

Key Features of NCERT Solutions for Class 10 Maths Chapter 14 – Statistics

• The solutions for the statistics chapter works as a reference for the students.
• It will help students to score marks against the questions asked from this chapter.
• Students can prepare and do the revision for Chapter 14 with this source.
• The questions of statistics have been solved by subject experts.
• The content of the material is as per the CBSE Syllabus (2023-24) and guidelines.

Statistics can also be understood in a more effective way by using the other solutions which are provided at BYJU’S. The solutions are prepared to help students perform well in the CBSE exams.

• RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics

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Dropped Topics –  14.5 Graphical representation of cumulative frequency distribution

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Class 10 - NCERT Mathematics Solutions

Exercise 13.1.

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean and why ?

By formula,

Class mark = Upper limit + Lower limit 2 \dfrac{\text{Upper limit + Lower limit}}{2} 2 Upper limit + Lower limit ​

We will use direct method for solving the mean of the above data as the numerical values of x i and f i are small.

Mean = Σ f i x i Σ f i = 162 20 \dfrac{Σf_ix_i}{Σf_i} = \dfrac{162}{20} Σ f i ​ Σ f i ​ x i ​ ​ = 20 162 ​ = 8.1

Hence, mean number of plants in each house = 8.1

Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Mean = Σ f i x i Σ f i = 27260 50 \dfrac{Σf_ix_i}{Σf_i} = \dfrac{27260}{50} Σ f i ​ Σ f i ​ x i ​ ​ = 50 27260 ​ = 545.20

Hence, mean daily wage = ₹ 545.20

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Mean = Σ f i x i Σ f i \dfrac{Σf_ix_i}{Σf_i} Σ f i ​ Σ f i ​ x i ​ ​

Substituting values we get :

⇒ 18 = 752 + 20 f 44 + f ⇒ 18 ( 44 + f ) = 752 + 20 f ⇒ 792 + 18 f = 752 + 20 f ⇒ 20 f − 18 f = 792 − 752 ⇒ 2 f = 40 ⇒ f = 40 2 = 20. \Rightarrow 18 = \dfrac{752 + 20f}{44 + f} \\[1em] \Rightarrow 18(44 + f) = 752 + 20f \\[1em] \Rightarrow 792 + 18f = 752 + 20f \\[1em] \Rightarrow 20f - 18f = 792 - 752 \\[1em] \Rightarrow 2f = 40 \\[1em] \Rightarrow f = \dfrac{40}{2} = 20. ⇒ 18 = 44 + f 752 + 20 f ​ ⇒ 18 ( 44 + f ) = 752 + 20 f ⇒ 792 + 18 f = 752 + 20 f ⇒ 20 f − 18 f = 792 − 752 ⇒ 2 f = 40 ⇒ f = 2 40 ​ = 20.

Hence, f = 20.

Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

We will use step deviation method.

In the following table a is the assumed mean and h is the class size.

Here, h = 3.

Mean = a + Σ f i u i Σ f i × h \dfrac{Σf_iu_i}{Σf_i} \times h Σ f i ​ Σ f i ​ u i ​ ​ × h

Mean  = 75.5 + 4 30 × 3 = 75.5 + 4 10 = 75.5 + 0.4 = 75.9 \text{Mean } = 75.5 + \dfrac{4}{30} \times 3 \\[1em] = 75.5 + \dfrac{4}{10} \\[1em] = 75.5 + 0.4 \\[1em] = 75.9 Mean  = 75.5 + 30 4 ​ × 3 = 75.5 + 10 4 ​ = 75.5 + 0.4 = 75.9

Hence, mean = 75.9

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ?

Here the given data is discontinuous.

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= 53 − 52 2 = 1 2 \dfrac{53 - 52}{2} = \dfrac{1}{2} 2 53 − 52 ​ = 2 1 ​ = 0.5

∴ 0.5 has to be added to the upper-class limit and 0.5 has to be subtracted from the lower-class limit of each interval.

We will use step deviation method to find the mean.

Mean  = 57 + 25 400 × 3 = 57 + 3 16 = 57 + 0.19 = 57.19 \text{Mean } = 57 + \dfrac{25}{400} \times 3 \\[1em] = 57 + \dfrac{3}{16} \\[1em] = 57 + 0.19 \\[1em] = 57.19 Mean  = 57 + 400 25 ​ × 3 = 57 + 16 3 ​ = 57 + 0.19 = 57.19

Hence, mean number of mangoes = 57.19

The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Here, h = 50.

Mean  = 225 + − 7 25 × 50 = 225 + ( − 14 ) = 211 \text{Mean } = 225 + \dfrac{-7}{25} \times 50 \\[1em] = 225 + (-14) \\[1em] = 211 Mean  = 225 + 25 − 7 ​ × 50 = 225 + ( − 14 ) = 211

Hence, mean daily expenditure on food = ₹ 211.

To find out the concentration of SO 2 (in parts per million, i.e., ppm), in the air, the data was collected for 30 localities in a certain city and is presented below :

Find the mean concentration of SO 2 in the air.

Here, h = 0.04

Mean  = 0.10 + − 1 30 × 0.04 = 0.10 − 0.04 30 = 0.10 − 4 3000 = 0.10 − 1 750 = 0.10 − 0.00133 = 0.099 \text{Mean } = 0.10 + \dfrac{-1}{30} \times 0.04 \\[1em] = 0.10 - \dfrac{0.04}{30} \\[1em] = 0.10 - \dfrac{4}{3000} \\[1em] = 0.10 - \dfrac{1}{750} \\[1em] = 0.10 - 0.00133 \\[1em] = 0.099 Mean  = 0.10 + 30 − 1 ​ × 0.04 = 0.10 − 30 0.04 ​ = 0.10 − 3000 4 ​ = 0.10 − 750 1 ​ = 0.10 − 0.00133 = 0.099

Hence, mean concentration of SO 2 = 0.099 ppm.

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

We will use assumed mean method to find the mean. Here, a is the assumed mean.

Mean = a + Σ f i d i Σ f i a + \dfrac{Σf_id_i}{Σf_i} a + Σ f i ​ Σ f i ​ d i ​ ​

Mean  = 17 + − 181 40 = 17 − 181 40 = 17 − 4.525 = 12.475 ≈ 12.48 \text{Mean } = 17 + \dfrac{-181}{40} \\[1em] = 17 - \dfrac{181}{40} \\[1em] = 17 - 4.525 \\[1em] = 12.475 ≈ 12.48 Mean  = 17 + 40 − 181 ​ = 17 − 40 181 ​ = 17 − 4.525 = 12.475 ≈ 12.48

Hence, mean number of days = 12.48

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Mean  = 70 + − 20 35 = 70 − 20 35 = 70 − 0.57 = 69.43 \text{Mean } = 70 + \dfrac{-20}{35} \\[1em] = 70 - \dfrac{20}{35} \\[1em] = 70 - 0.57 \\[1em] = 69.43 Mean  = 70 + 35 − 20 ​ = 70 − 35 20 ​ = 70 − 0.57 = 69.43

Hence, mean literacy rate = 69.43 %.

Exercise 13.2

The following table shows the ages of the patients admitted in a hospital during a year :

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

We will find mean by direct method.

Mean = Σ f i x i Σ f i = 2830 80 = 35.375 \dfrac{Σf_ix_i}{Σf_i} = \dfrac{2830}{80} = 35.375 Σ f i ​ Σ f i ​ x i ​ ​ = 80 2830 ​ = 35.375

Mode = l + ( f 1 − f 0 2 f 1 − f 0 − f 2 ) × h \Big(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\Big) \times h ( 2 f 1 ​ − f 0 ​ − f 2 ​ f 1 ​ − f 0 ​ ​ ) × h

Class size is h.

The lower limit of modal class is l

The Frequency of modal class is f 1 .

Frequency of class preceding modal class is f 0 .

Frequency of class succeeding the modal class is f 2 .

Class 35 - 45 has the highest frequency.

∴ It is the modal class.

∴ l = 35, f 1 = 23, f 0 = 21, f 2 = 14 and h = 10.

Mode = 35 + ( 23 − 21 2 × 23 − 21 − 14 ) × 10 = 35 + 2 46 − 35 × 10 = 35 + 20 11 = 35 + 1.8 = 36.8 \text{Mode} = 35 + \Big(\dfrac{23 - 21}{2 \times 23 - 21 - 14}\Big) \times 10 \\[1em] = 35 + \dfrac{2}{46 - 35} \times 10 \\[1em] = 35 + \dfrac{20}{11} \\[1em] = 35 + 1.8 \\[1em] = 36.8 Mode = 35 + ( 2 × 23 − 21 − 14 23 − 21 ​ ) × 10 = 35 + 46 − 35 2 ​ × 10 = 35 + 11 20 ​ = 35 + 1.8 = 36.8

Since, mode = 36.8

∴ Maximum number of patients admitted in the hospital are of the age 36.8 years.

Since, mean = 35.37

∴ Average the age of a patient admitted to the hospital is 35.37 years.

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Class 60 - 80 has the highest frequency.

∴ l = 60, f 1 = 61, f 0 = 52, f 2 = 38 and h = 20.

Mode = 60 + ( 61 − 52 2 × 61 − 52 − 38 ) × 20 = 60 + 9 122 − 90 × 20 = 60 + 180 32 = 60 + 5.625 = 65.625 \text{Mode} = 60 + \Big(\dfrac{61 - 52}{2 \times 61 - 52 - 38}\Big) \times 20 \\[1em] = 60 + \dfrac{9}{122 - 90} \times 20 \\[1em] = 60 + \dfrac{180}{32} \\[1em] = 60 + 5.625 \\[1em] = 65.625 Mode = 60 + ( 2 × 61 − 52 − 38 61 − 52 ​ ) × 20 = 60 + 122 − 90 9 ​ × 20 = 60 + 32 180 ​ = 60 + 5.625 = 65.625

Hence, modal lifetime = 65.625 hours.

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Class 1500 - 2000 has the highest frequency.

∴ l = 1500, f 1 = 40, f 0 = 24, f 2 = 33 and h = 500.

Mode = 1500 + ( 40 − 24 2 × 40 − 24 − 33 ) × 500 = 1500 + 16 80 − 57 × 500 = 1500 + 16 23 × 500 = 1500 + 8000 23 = 1500 + 347.83 = 1847.83 \text{Mode} = 1500 + \Big(\dfrac{40 - 24}{2 \times 40 - 24 - 33}\Big) \times 500 \\[1em] = 1500 + \dfrac{16}{80 - 57} \times 500 \\[1em] = 1500 + \dfrac{16}{23} \times 500 \\[1em] = 1500 + \dfrac{8000}{23} \\[1em] = 1500 + 347.83 \\[1em] = 1847.83 Mode = 1500 + ( 2 × 40 − 24 − 33 40 − 24 ​ ) × 500 = 1500 + 80 − 57 16 ​ × 500 = 1500 + 23 16 ​ × 500 = 1500 + 23 8000 ​ = 1500 + 347.83 = 1847.83

We will find mean using step deviation method.

Mean  = 3250 + − 235 200 × 500 = 3250 − 1175 2 = 3250 − 587.5 = 2662.50 \text{Mean } = 3250 + \dfrac{-235}{200} \times 500 \\[1em] = 3250 - \dfrac{1175}{2} \\[1em] = 3250 - 587.5 \\[1em] = 2662.50 Mean  = 3250 + 200 − 235 ​ × 500 = 3250 − 2 1175 ​ = 3250 − 587.5 = 2662.50

Hence, mean = ₹ 2662.50 and mode = ₹ 1847.83.

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Class 30 - 35 has the highest frequency.

∴ l = 30, f 1 = 10, f 0 = 9, f 2 = 3 and h = 5.

Mode = 30 + ( 10 − 9 2 × 10 − 9 − 3 ) × 5 = 30 + 1 20 − 12 × 5 = 30 + 5 8 = 30 + 0.625 = 30.625 ≈ 30.6 \text{Mode} = 30 + \Big(\dfrac{10 - 9}{2 \times 10 - 9 - 3}\Big) \times 5 \\[1em] = 30 + \dfrac{1}{20 - 12} \times 5 \\[1em] = 30 + \dfrac{5}{8} \\[1em] = 30 + 0.625 \\[1em] = 30.625 ≈ 30.6 Mode = 30 + ( 2 × 10 − 9 − 3 10 − 9 ​ ) × 5 = 30 + 20 − 12 1 ​ × 5 = 30 + 8 5 ​ = 30 + 0.625 = 30.625 ≈ 30.6

We will find mean using direct method.

Mean = Σ f i x i Σ f i = 1022.5 35 \dfrac{Σf_ix_i}{Σf_i} = \dfrac{1022.5}{35} Σ f i ​ Σ f i ​ x i ​ ​ = 35 1022.5 ​ = 29.2

Hence, mode = 30.6 and median = 29.2 so, most states/U.T. have a student teacher ratio of 30.6 and on an average, this ratio is 29.2

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Class 4000 - 5000 has the highest frequency.

∴ l = 4000, f 1 = 18, f 0 = 4, f 2 = 9 and h = 1000.

Mode = 4000 + ( 18 − 4 2 × 18 − 4 − 9 ) × 1000 = 4000 + 14 36 − 13 × 1000 = 4000 + 14000 23 = 4000 + 608.7 = 4608.7 \text{Mode} = 4000 + \Big(\dfrac{18 - 4}{2 \times 18 - 4 - 9}\Big) \times 1000 \\[1em] = 4000 + \dfrac{14}{36 - 13} \times 1000 \\[1em] = 4000 + \dfrac{14000}{23} \\[1em] = 4000 + 608.7 \\[1em] = 4608.7 Mode = 4000 + ( 2 × 18 − 4 − 9 18 − 4 ​ ) × 1000 = 4000 + 36 − 13 14 ​ × 1000 = 4000 + 23 14000 ​ = 4000 + 608.7 = 4608.7

Hence, mode = 4608.7 runs.

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Class 40 - 50 has the highest frequency.

∴ l = 40, f 1 = 20, f 0 = 12, f 2 = 11 and h = 10.

Mode = 40 + ( 20 − 12 2 × 20 − 12 − 11 ) × 10 = 40 + 8 40 − 23 × 10 = 40 + 80 17 = 40 + 4.7 = 44.7 \text{Mode} = 40 + \Big(\dfrac{20 - 12}{2 \times 20 - 12 - 11}\Big) \times 10 \\[1em] = 40 + \dfrac{8}{40 - 23} \times 10 \\[1em] = 40 + \dfrac{80}{17} \\[1em] = 40 + 4.7 \\[1em] = 44.7 Mode = 40 + ( 2 × 20 − 12 − 11 20 − 12 ​ ) × 10 = 40 + 40 − 23 8 ​ × 10 = 40 + 17 80 ​ = 40 + 4.7 = 44.7

Hence, mode = 44.7 cars.

Exercise 13.3

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

We will find mean by step deviation method.

Class mark = Lower limit + Upper limit 2 \dfrac{\text{Lower limit + Upper limit}}{2} 2 Lower limit + Upper limit ​

Here, h (class size) = 20.

Mean = a + Σ f i u i Σ f i × h a + \dfrac{Σf_iu_i}{Σf_i} \times h a + Σ f i ​ Σ f i ​ u i ​ ​ × h

Mean  = 135 + 7 68 × 20 = 135 + 35 17 = 135 + 2.05 = 137.05 \text{Mean } = 135 + \dfrac{7}{68} \times 20 \\[1em] = 135 + \dfrac{35}{17} \\[1em] = 135 + 2.05 = 137.05 Mean  = 135 + 68 7 ​ × 20 = 135 + 17 35 ​ = 135 + 2.05 = 137.05

Cumulative frequency distribution table is as follows :

Here, n = 68, which is even

n 2 = 68 2 \dfrac{n}{2} = \dfrac{68}{2} 2 n ​ = 2 68 ​ = 34.

Cumulative frequency just greater than n 2 \dfrac{n}{2} 2 n ​ is 42, belonging to class-interval 125 - 145.

∴ Median class = 125 - 145

⇒ Lower limit of median class (l) = 125

⇒ Frequency of median class (f) = 20

⇒ Cumulative frequency of class preceding median class (cf) = 22

Median = l + ( n 2 − c f f ) × h l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h l + ( f 2 n ​ − c f ​ ) × h

Median = 125 + 34 − 22 20 × 20 = 125 + 12 20 × 20 = 125 + 12 = 137. \text{Median} = 125 + \dfrac{34 - 22}{20} \times 20 \\[1em] = 125 + \dfrac{12}{20} \times 20 \\[1em] = 125 + 12 \\[1em] = 137. Median = 125 + 20 34 − 22 ​ × 20 = 125 + 20 12 ​ × 20 = 125 + 12 = 137.

From table,

Class 125 - 145 has the highest frequency.

∴ l = 125, f 1 = 20, f 0 = 13, f 2 = 14 and h = 20.

Mode = 125 + ( 20 − 13 2 × 20 − 13 − 14 ) × 20 = 125 + 7 40 − 27 × 20 = 125 + 140 13 = 125 + 10.76 = 135.76 \text{Mode} = 125 + \Big(\dfrac{20 - 13}{2 \times 20 - 13 - 14}\Big) \times 20 \\[1em] = 125 + \dfrac{7}{40 - 27} \times 20 \\[1em] = 125 + \dfrac{140}{13} \\[1em] = 125 + 10.76 \\[1em] = 135.76 Mode = 125 + ( 2 × 20 − 13 − 14 20 − 13 ​ ) × 20 = 125 + 40 − 27 7 ​ × 20 = 125 + 13 140 ​ = 125 + 10.76 = 135.76

Hence, mean = 137.05, median = 137 and mode = 135.76

If the median of the distribution given below is 28.5, find the values of x and y.

We know that,

⇒ 45 + x + y = 60

⇒ x + y = 15 ...........(1)

Median = 28.5

From cumulative frequency distribution table we get :

Median lies in class 20 - 30.

∴ Median class = 20 - 30

⇒ Lower limit of median class (l) = 20

⇒ Class size (h) = 10

⇒ Cumulative frequency of class preceding median class (cf) = 5 + x

⇒ 28.5 = 20 + ( 60 2 − ( 5 + x ) 20 ) × 10 ⇒ 28.5 − 20 = 30 − 5 − x 2 ⇒ 8.5 × 2 = 25 − x ⇒ 17 = 25 − x ⇒ x = 25 − 17 = 8. \Rightarrow 28.5 = 20 + \Big(\dfrac{\dfrac{60}{2} - (5 + x)}{20}\Big) \times 10 \\[1em] \Rightarrow 28.5 - 20 = \dfrac{30 - 5 - x}{2} \\[1em] \Rightarrow 8.5 \times 2 = 25 - x \\[1em] \Rightarrow 17 = 25 - x \\[1em] \Rightarrow x = 25 - 17 = 8. ⇒ 28.5 = 20 + ( 20 2 60 ​ − ( 5 + x ) ​ ) × 10 ⇒ 28.5 − 20 = 2 30 − 5 − x ​ ⇒ 8.5 × 2 = 25 − x ⇒ 17 = 25 − x ⇒ x = 25 − 17 = 8.

Substituting value of x in equation (1), we get :

⇒ 8 + y = 15

⇒ y = 15 - 8

Hence, x = 8 and y = 7.

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Here, n = 100, n 2 = 50 \dfrac{n}{2} = 50 2 n ​ = 50 .

Cumulative frequency just greater than 50 is 78, belonging to class-interval 35 − 40.

Therefore, median class = 35 - 40

⇒ Class size (h) = 5

⇒ Lower limit of median class (l) = 35

⇒ Frequency of median class (f) = 33

⇒ Cumulative frequency of class preceding median class (cf) = 45

⇒ Median  = 35 + 50 − 45 33 × 5 = 35 + 25 33 = 35 + 0.76 = 35.76 \Rightarrow \text{Median } = 35 + \dfrac{50 - 45}{33} \times 5 \\[1em] = 35 + \dfrac{25}{33} \\[1em] = 35 + 0.76 \\[1em] = 35.76 ⇒ Median  = 35 + 33 50 − 45 ​ × 5 = 35 + 33 25 ​ = 35 + 0.76 = 35.76

Hence, median age = 35.76 years.

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Find the median length of leaves.

= 127 − 126 2 = 1 2 \dfrac{127 - 126}{2} = \dfrac{1}{2} 2 127 − 126 ​ = 2 1 ​ = 0.5

Here, n = 40, n 2 = 40 2 = 20 \dfrac{n}{2} = \dfrac{40}{2} = 20 2 n ​ = 2 40 ​ = 20 .

Cumulative frequency just greater than 20 is 29, belonging to class-interval 144.5 − 153.5

Therefore, median class = 144.5 - 153.5

⇒ Class size (h) = 9

⇒ Lower limit of median class (l) = 144.5

⇒ Frequency of median class (f) = 12

⇒ Cumulative frequency of class preceding median class (cf) = 17

⇒ Median  = 144.5 + 20 − 17 12 × 9 = 144.5 + 27 12 = 144.5 + 2.25 = 146.75 \Rightarrow \text{Median } = 144.5 + \dfrac{20 - 17}{12} \times 9 \\[1em] = 144.5 + \dfrac{27}{12} \\[1em] = 144.5 + 2.25 \\[1em] = 146.75 ⇒ Median  = 144.5 + 12 20 − 17 ​ × 9 = 144.5 + 12 27 ​ = 144.5 + 2.25 = 146.75

Hence, median length = 146.75 mm.

The following table gives the distribution of the life time of 400 neon lamps :

Find the median life time of a lamp.

Here, n = 400, n 2 = 400 2 = 200 \dfrac{n}{2} = \dfrac{400}{2} = 200 2 n ​ = 2 400 ​ = 200 .

Cumulative frequency just greater than 200 is 216, belonging to class-interval 3000 - 3500

∴ Median class = 3000 - 3500

⇒ Class size (h) = 500

⇒ Lower limit of median class (l) = 3000

⇒ Frequency of median class (f) = 86

⇒ Cumulative frequency of class preceding median class (cf) = 130

⇒ Median  = 3000 + 200 − 130 86 × 500 = 3000 + 35000 86 = 3000 + 406.98 = 3406.98 \Rightarrow \text{Median } = 3000 + \dfrac{200 - 130}{86} \times 500 \\[1em] = 3000 + \dfrac{35000}{86} \\[1em] = 3000 + 406.98 \\[1em] = 3406.98 ⇒ Median  = 3000 + 86 200 − 130 ​ × 500 = 3000 + 86 35000 ​ = 3000 + 406.98 = 3406.98

Hence, median life = 3406.98 hours.

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

We will use direct method to find the mean.

Mean = Σ f i x i Σ f i = 832 100 \dfrac{Σf_ix_i}{Σf_i} = \dfrac{832}{100} Σ f i ​ Σ f i ​ x i ​ ​ = 100 832 ​ = 8.32

Here, n = 100, n 2 = 100 2 \dfrac{n}{2} = \dfrac{100}{2} 2 n ​ = 2 100 ​ = 50.

Cumulative frequency just greater than 50 is 76, belonging to class-interval 7 - 10.

∴ Median class = 7 - 10

⇒ Class size (h) = 3

⇒ Lower limit of median class (l) = 7

⇒ Frequency of median class (f) = 40

⇒ Cumulative frequency of class preceding median class (cf) = 36

Median = 7 + 50 − 36 40 × 3 = 7 + 14 × 3 40 = 7 + 42 40 = 7 + 1.05 = 8.05 \text{Median} = 7 + \dfrac{50 - 36}{40} \times 3 \\[1em] = 7 + \dfrac{14 \times 3}{40} \\[1em] = 7 + \dfrac{42}{40} \\[1em] = 7 + 1.05 \\[1em] = 8.05 Median = 7 + 40 50 − 36 ​ × 3 = 7 + 40 14 × 3 ​ = 7 + 40 42 ​ = 7 + 1.05 = 8.05

Class 7 - 10 has the highest frequency.

∴ l = 7, f 1 = 40, f 0 = 30, f 2 = 16 and h = 3.

Mode = 7 + ( 40 − 30 2 × 40 − 30 − 16 ) × 3 = 7 + 10 80 − 46 × 3 = 7 + 30 34 = 7 + 0.88 = 7.88 \text{Mode} = 7 + \Big(\dfrac{40 - 30}{2 \times 40 - 30 - 16}\Big) \times 3 \\[1em] = 7 + \dfrac{10}{80 - 46} \times 3 \\[1em] = 7 + \dfrac{30}{34} \\[1em] = 7 + 0.88 \\[1em] = 7.88 Mode = 7 + ( 2 × 40 − 30 − 16 40 − 30 ​ ) × 3 = 7 + 80 − 46 10 ​ × 3 = 7 + 34 30 ​ = 7 + 0.88 = 7.88

Hence, mean = 8.32, median = 8.05 and mode = 7.88

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Here, n = 30, n 2 = 30 2 = 15 \dfrac{n}{2} = \dfrac{30}{2} = 15 2 n ​ = 2 30 ​ = 15 .

Cumulative frequency just greater than 15 is 19, belonging to class-interval 55 - 60

∴ Median class = 55 - 60

Class size (h) = 5

Lower limit of median class (l) = 55

Frequency of median class (f) = 6

Cumulative frequency of class preceding median class (cf) = 13

⇒ Median  = 55 + 15 − 13 6 × 5 = 55 + 10 6 = 55 + 1.67 = 56.67 \Rightarrow \text{Median } = 55 + \dfrac{15 - 13}{6} \times 5 \\[1em] = 55 + \dfrac{10}{6} \\[1em] = 55 + 1.67 \\[1em] = 56.67 ⇒ Median  = 55 + 6 15 − 13 ​ × 5 = 55 + 6 10 ​ = 55 + 1.67 = 56.67

Hence, median weight = 56.67 kg.

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CBSE Class 10 Maths Case Study Questions for Chapter 14 - Statistics (Published By CBSE)

Case study question bank for cbse class 10 maths chapter 14 - statistics is available here. practice this new format of questions to score good marks in your board exam..

CBSE Class 10 Maths Case Study Questions for Chapter 14 - Statistics are published by the CBSE board itself. These questions are perfect to acquaint with the new format of the questions and make your board exam preparations. Questions are based on the real life situations. You can easily understand how concepts and logic are used in the case study questions. All the questions are provided with answers.  

Check Case Study Questions for Class 10 Maths Chapter 14 - Statistics

CASE STUDY 1:

COVID-19 Pandemic The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans.

The following tables shows the age distribution of case admitted during a day in two different hospitals

Refer to table 1

1. The average age for which maximum cases occurred is

Answer: c) 36.82

2. The upper limit of modal class is

Answer: d) 45

3. The mean of the given data is

Answer: d) 35.4

Refer to table 2

4. The mode of the given data is

Answer: a) 41.4

5. The median of the given data is

Answer: b) 40.2

Electricity Energy Consumption

CASE STUDY 2:

Electricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption).

Refer to data received from Colony A

1. The median weekly consumption is

a) 12 units

b) 16 units

c) 20 units

d) None of these

Answer: c) 20 units

2. The mean weekly consumption is

a) 19.64 units

b) 22.5 units

c) 26 units

Answer: a) 19.64 units

3. The modal class of the above data is I

Answer: c) 20-30

Refer to data received from Colony B

4. The modal weekly consumption is

a) 38.2 units

b) 43.6 units

d) 32 units

Answer: b) 43.6 units

5. The mean weekly consumption is

a) 15.65 units

b) 32.8 units

c) 38.75 units

d) 48 units

Answer: c) 38.75 units

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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NCERT Revision Notes for Chapter 14 Statistics Class 10 Maths

Average or mean is the certain value representative of the whole data and signifying its characteristics is called average of the data. It is also known as mean.

The average is given as

Median of data is central data, obtained when data is arranged in ascending order. Mode is the data, having maximum frequency. 

Grouped Frequency Table:

 Presenting a large number of data, we need to group data for certain range, and get the frequency of that group. 

 Presenting a large number of data, we need to group data for certain range, and get the frequency of that group.

 Here Frequency of class Size is 0-4 is 3. It means 3 students has obtained marks in the range of 0-4. 

Above is the grouped the data, having range of 4 marks, i.e 0-4, 4-9.These groupings are called ‘classes’ or ‘class-intervals’, and their size is called the class-size or class width.

Class width is chosen according convenience. 

Consider the following Grouped Data

 In class size, if there is 10 as one of the observation, it is not represented in 0-10. It is represented in 10-20. Similarly, If 20 is one of the data; it is included in 20-30. Not in 10-20.  So, upper limit is included, and lower limit is excluded.

Inclusive Grouped Data

In this, both upper limit and lower limit is included.

 That is in class 0-9, both 9 is also included.

Mean of an Inclusive Series

To find mean of inclusive series, it is first needed to be converted into exclusive series.

It is done, by subtracting 0.5 from upper and adding 0.5 to lower limit. After that, you can calculate mean by whatever method you like, or by the method which is mentioned in question. 

The class size will be 1 more for exclusive series for corresponding inclusive series. 

Note:- After converting Inclusive series in exclusive series, we can find mean, median and mode of the given data. 

So, here we have inclusive series, We will convert it into exclusive like below 

Cumulative frequency is the running total of all frequencies.

Consider the following Data:-

 The following data represent the marks obtained by 100 students in science test.

 As you can see that, cumulative frequency is the sum of all frequencies of all class size, till that class.

What Cumulative frequency signifies?

 So, by using cumulative frequency, we get that, how many students scored less than a particular mark. Similarly, we can represent the above data, in following form. 

More than and Less Than

Tabulate the ‘less than’ and ‘more than’ cumulative series, as mentioned earlier. 

On a graph paper, we mark the upper class limits along X axis and corresponding cumulative frequencies along y axis.

Take a point A(0,N/2) on the y axis and draw AP II X-axis, cutting the above curve at a point P. Draw PM perpendicular to x-axis.

We can draw it for both greater than and lower than series.

The point as much both curve intersect is median.

The curve obtained by less than series is known as less than ogive. The curve obtained by more than series is known as more than Ogive.

Q. Draw less than and more than Ogive of the following data, and find the median.

Now, we will draw a more than series table of the above data

Now, we will draw less than series.

Now, we will plot Graphs for both, one by one.

NCERT Solutions for Chapter 4 The Age of Industrialisation Class 10 History

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• Pair of Linear Equations in Two Variables
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Important Questions for Class 10 Maths Chapter 14 Statistics

August 9, 2019 by Sastry CBSE

Statistics Class 10 Important Questions Very Short Answer (1 Mark)

Question 1. In a continuous frequency distribution, the median of the data is 21. If each observation is increased by 5, then find the new median. (2015) Solution: New median = 21 + 5 = 26

Statistics Class 10 Important Questions Short Answer-I (2 Marks)

Question 13. Find the median of the data using an empirical formula, when it is given that mode = 35.3 and mean = 30.5. (2014) Solution: Mode = 3(Median) – 2(Mean) 35.3 = 3(Median) – 2(30.5) 35.3 = 3(Median) – 61 96.3 = 3 Median Median = \(\frac{96.3}{3}\) = 32.1

Question 14. Show that the mode of the series obtained by combining the two series S 1 and S 2 given below is different from that of S 1 and S 2 taken separately: (2015) S 1 : 3, 5, 8, 8, 9, 12, 13, 9, 9 S 2 : 7, 4, 7, 8, 7, 8, 13 Solution: In S 1 : Number 9 occurs 3 times (maximum) ∴ Mode of S 1 Series = 9 In S 2 : Number 7 occurs 3 times (maximum) ∴ Mode of S, Series = 7 After combination: In S 1 & S 2 : No. 8 occurs 4 times (maximum) ∴ Mode of S 1 & S 2 taken combined = 8 So, mode of S 1 & S 2 combined is different from that of S 1 & S 2 taken separately.

Statistics Class 10 Important Questions Short Answer-II (3 Marks)

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• All Mathematics assignments for Class 10 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
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• Class 10 Mathematics question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 10 Mathematics students

At https://www.cbsencertsolutions.com, we have provided the biggest database of free assignments for Mathematics Class 10 which you can download in Pdf

We provide here Standard 10 Mathematics chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Mathematics in Grade 10, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

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Assignments For Class 10 Mathematics Statistics

Assignments for Class 10 Mathematics Statistics have been developed for Standard 10 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 10 Mathematics Statistics from our website as we have provided all Maths Assignment for Class 10 chapter wise free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 10 Mathematics Statistics. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Mathematics Statistics book and get good marks in class 10 exams.

Question . The median and mode of a frequency distribution are 26 and 29 respectively. Then, the mean is  (A) 27.5 (B) 24.5 (C) 28.4 (D) 25.8

Question . If the mean and median of a set of numbers are 8.9 and 9 respectively, then the mode will be    (A) 7.2 (B) 8.2 (C) 9.2 (D) 10.2

Question . Look at the frequency distribution table given below:      The median of the above distribution is

(A) 56.5 (B) 57.5 (C) 58.5 (D) 59

Question . Mode = ?       

Question . For a symmetrical frequency distribution, we have    (A) mean < mode < median (B) mean > mode > median (C) mean = mode = median (D) mode = 1/2 (mean median)

Question . Look at the cumulative frequency distribution table given below:    Number of families having income range 20000 to 25000 is

(A) 19 (B) 16 (C) 13 (D) 22

Measure of Central Tendency Mathematical Calculations

Where,l= lower limit of the model class, h=size of the class-interval,

Where, lower limit of the model class, size of the class-interval, f 1 = frequency of the model class,f 2 =frequency of the class, preceding the modal class,

f 2 =frequency the modal class,h=class size.

Where, l=lower limit of the median class n=number of observations, c.f=cumulative frequency if the class preceding the median class, f=frequency of the median class, h=class size.

Graphical Representation Cumulative Frequency Graph Less than Ogive More then Ogive All the formulae discussed are for grouped data calculation for ungrouped data have been done in previous class The three measures mean, mode and median are connected by the following relations. Mode = 3 median – 2 mean

Short / Long Answer types Questions :

Question. If the mean of the following data is 20.6, find the value of p.

Solution.      

Question. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ` 18. Find the missing frequency k. 

Solution.   k = 8

Question. Find the mean of the following data:   

Solution.    

Question. The mean of the following frequency distribution is 62.8 and sum of all frequencies is 50. Find the missing frequencies f 1 and f 2 .

Question. The table below shows the daily expenditure on grocery of 25 households in a locality.   

Find the mean daily expenditure on food by a suitable method. Solution.    

Question. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ` 18. Find the missing frequency f. 

Question. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. 

Question. Find the class-mark of class 25–35. Solution.   Class-mark = 25 + 35/2 = 60/2 = 30

Question. Find the mean of first ten odd natural numbers. Solution.   10

Question. If the mean of the first n natural number is 15, then find n.    Solution.   15 = 1 + 2 + 3 + … + n/2

Question. Find the mean of the following distribution:

Solution.   

Question. Find the class-marks of the classes 10-25 and 35-55. Solution.   Class-mark of class 10 – 25  = 10 + 25/2 = 35/2 = 17.5 Class-mark of class 35–55 = 35 + 55/2 = 90/2 = 45

Question. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are as follows:   

Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km / l. Do you agree with this claim? Solution. 

Hence, mean mileage of car is 14.48 km/litre. So, the manufacturer’s statement is wrong that mileage is 16 km L –1 .     

Question. The arithmetic mean of the following frequency distribution is 53. Find the value of k.   

⇒ 3340 + 70k = 53 (72 + k) ⇒ 3340 + 70k = 3816 + 53k ⇒ 70k – 53k = 3816 – 3340 ⇒ 17k = 476 ⇒ k = 28

Question. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given as follows: 

Determine the mean number of seats occupied over the flights. Solution.    

Question. If the mean of the following distribution is 6.4, then find the value of ‘p’.

Solution. 

Question. Find the arithmetic mean of 1, 2, 3, …, n          Solution.  We know that,   

Question. The following data gives the information on the observed life-times (in hours) of 225 electrical components.   

Determine the mean of the above data. Solution.

PRACTICE EXERCISE

Question. Calculate the median for the following data :   

Solution.  153.4

Question. Draw the ogive for the following frequency distribution :   

Also, find the median using ogive drawn. Solution.  37.5

Question. Find the mean of the following frequency distribution :   

Solution.  48.70

Question. Calculate the mode of the following data :   

Solution.  77.27

Question. Calculate the arithmetic mean of the following frequency distribution : 

Solution.  69.33

Question. The mean of the following frequency table is 50. But the frequencies f 1  and f 2  in class 20-40 and 60-80 are missing. Find the missing frequencies.   

Solution.  f 1  = 28, f 2  = 24

Question. The following table shows the weights in gm of a sample of 100 potatoes taken from a large consignment:   

Draw the cumulative frequency curve and from it determine the median weight of the potatoes. Solution.  93 gm

Question. The marks of 200 students in a test were recorded as follows : 

Draw the cumulative frequency table and ogive. Use it to estimate the median. Solution.  52.8

Question. The following table gives the distribution of expenditure of different families on education. Find the mean expenditure on education in a family. 

Solution.  Rs. 2662.5

Question. Calculate the modal height from the following table   

Solution.  153.57 cm

Question. Find the value of p if the mean of the following distribution is 7.5.   

Solution.  p = 3

Question. Find the mode of the following distribution :     

Solution.  52

Solution.  25.2

Question. Find the average marks scored by students of a class in a test from the following data : 

Solution.  28 marks

Question. For the following frequency distribution, draw both the types of cumulative frequency curve on the same graph paper and hence find the median.   

Solution.  75

Question. Draw a ‘less than type’ cumulative frequency curve for the following data. 

Hence, estimate the median. Solution.  20

Question. Draw a cumulative frequency curve (less than type) for the following data and find the median from it.   

Solution.  32.6

Question. Find the mean of the following distribution :   

Solution.  19.92

Question. Calculate the mode of the following distribution :     

Solution.  107

Question. The marks obtained by 100 students of a class in an examination are given below :

Draw a cumulative frequency curved by using : (i) ‘less than type’ and (ii) ‘more than type’. Hence find the meadian. Solution.  28.8

Question. Compute the mode of the following data :   

Solution.  85.71

Question. Calculate the mode from the following data :   

Solution.  Rs. 7727.27

Question. Marks scored by 400 students in an examination are as follows :

Draw a ‘more than type’ cumulative frequency curve and from it determine the median. Solution.  57.5

Question. Calculate the mean, the median and the mode of the following distribution :   

Solution.  Mean = 14.96, Median = 15, Mode = 15

Question. The table given below shows the distribution of the daily wages, earned by 160 workers in a building site: 

Draw a cumulative frequency curve by using : (i) ‘less than type’ and (ii) ‘more than type’. Hence, estimate the median wages using graph. Solution.  Rs. 34.73

Question. Find the median of the following frequency distribution : 

Solution.  27

Solution.  55

Question. Calculate the mode for the following frequency distribution : 

Solution.  52.83 kg

Question. Find the mean of the following frequency distribution : 

Solution.  53

Question. Calculate the mean of the following frequency distribution: 

Solution.  196.8

Question. The arithmetic mean of the following frequency distribution is 50. Find the value of p. 

Solution.  p = 28

Solution.  50

Question. Calculate the median income :   

Solution.  Rs. 934.17

Solution.  148.61

Question. Find the median for the following frequency distribution :   

Solution.  35.76 years (approx).

Question. Find the mode for the following frequency distribution :   

Solution.  63.75

Question. Find the average height for the following frequency distribution : 

Solution.  153.45 cm

Solution.  153.64

Question. The following table shows the ages of the patients admitted in a hospital during a year : 

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Solution.  mode = 36.81 years; mean = 35.37 years

Question. The following table gives the marks obtained by 50 students in a class test :   

Find the median marks. Solution.  33

Question. Calculate the mean of the following distribution :   

Solution.  62.545

Question. The median of the following distribution is 35, find the value of a and b.   

Solution.  a = 35, b = 25

Solution.  35

Question. Find the mode for the following distribution table :   

Solution.  44.705

Solution.  12.89

Solution.  116.67

Question. Draw a cumulative frequency curve (more than type) for the following data and hence obtain the median.   

Solution.  35 (approx)

Question. If the median of the distribution given below is 28.5, find the value of x and y. 

Solution.  x = 8, y = 7

Question. If the mean of the distribution is 57.6 and the sum of its observations is 50, find the missing frequencies f 1 and f 2 .   

Solution.  f 1 = 8, f 2  = 10

Solution.  36.357

Question. Compute the mean for the following data :   

Solution.  24

Question. If the mean of the following distribution is 54, find the value of p :

Solution.  p = 11

Question. The wickets taken by a bowler in10 cricket matches are as follows:2,6,4,5,0,2,1,3,2,3 Find the mode of the data. Solution.  2

Question. How one can find median of a frequency distribution graphically. Solution.  OGIVE

Question. What is the mean of first ten prime numbers? Solution.  12.9

Question. What measure of central tendency is represented by the abscissa of the point where less than ogive and more than ogive intersect? Solution.  MEDIAN

Question. What important information one can get by the abscissa of the point of intersection of the less than type and the more than type cumulative frequency curve of a group data. Solution.  Median

Question. Using the assumed mean method find the mean of the following data.     

Solution.  27.2

Question. If the mode of a data is 45 and mean is 27, then median is .   Solution.  33

Question. Find the median of the following frequency distribution.   

Solution.  167

Question. Given below is the distribution of IQ of the 100 students .Find the median of IQ   

Solution.  106.1

Question. Find the median of the following distribution.   

Solution.  28.5l

Question. A class teacher has the following absentee records of 40 students of a class for the whole term 

Write the above distribution as less than type cumulative frequency distribution Solution.   

Question. Find the mode of the following     

Solution.  MODE =40

Question. Name the key terms used in central tendency Solution.  Mean, Median, Mode

Question. Draw cumulative frequency curve for the following data :   

Hence, determine the median. Solution.  32.5

Solution.  43

Question. Find the mode of the following data:  120, 110, 130, 110, 120, 140, 130, 120, 140, 120  Solution.  Let us first form the frequency table for the given data as given below: 

We observe that the value 120 has the maximum frequency. Hence, the mode or modal values is 120.

Question. Calculate the average daily income (in Rs) of the following data about men working in a company:   

Find the median for the above frequency distribution.    Solution.  n = 120 ⇒ n/2 = 60 Median is average of 60th and 61st observation Median = 5 + 5 /2 = 5 So, Median is 5.

Question. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute was recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. 

Solution.  Take a = 75.5, h = 3

Using the step-deviation method,   

Hence, the mean heart beats per minute are 75.9.

Question. Calculate the median from the following data:   

Question. Find the unknown entries a, b, c, d in the following distribution of heights of students in a class :   

Question. On annual day of a school, 400 students participated in the function. Frequency distribution showing their ages is as shown in the following table : 

Find mean and median of the above data. Solution. 

a= assumed mean = 12 we know that,   

Question. The data regarding marks obtained by 48 students of a class in a class test is given below. Calculate the modal marks of students.   

Solution.  Modal class is 30 – 35, l =30, f 1 = 25, f 0  = 10, 

= 30 + 2.27 or 32.27 approx.   

Question. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f 1 and f 2 .   

Solution.  Given,    sum of frequency = 50 5 + f 1  + 10 + f 2  + 7 + 8 = 50 f 1  + f 2  = 20 3f 1  + 3f 2  = 60 … (1) [multiply both side by 3] And mean = 62.8

30f 1  + 70f 2  = 3140 – 2060 30f 1  + 70f 2  = 1080 3f 1  + 7f 2  = 108 …(2) [divide it by 10] subtract equation (1) from equation (2) 3f 1  + 7f 2  – 3f 1  – 3f 2  =108 – 60 4f 2  = 48 f 2  =12 Put value of f 2  in equation (1) 3f 1  + 3(12) = 60 f 1  = 24/3 = 8 f 1  = 8, f 2  = 12

Question. From the following frequency distribution, find the median class : 

Question. The arithmetic mean of the following data is 14. Find the value of k. 

Question. Find the mode of the data , using a empirical formula , when it is given that median = 41.25 and mean = 33.75   Solution.  We know , Mode = 3(Median) – 2(Mean) Given, Median = 41.25 and Mean = 33.75 Therefore , Mode = 3 x 41.25 – 2 x 33.75 = 123. 75 – 67.50 = 56.25 So, mode is 56.25.

Question. The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs.18.00. Find out the missing frequency. 

Solution.  Given mean = 18, Let the missing frequency be ‘v’. 

Question. The following table gives the number of children of 150 families in a village 

Find the average number of children per family.    Solution.  Let the assumed mean (A) = 2 

Average number of children per family = 352/150 = 2.35(approx)

Question. Find the mode of the following data:  25,16,19, 48,19, 20,34,15, 19, 20, 21, 24,19, 16, 22, 16 ,18, 20,16, 19 ……. Solution.  Given, data 25,16,19, 48,19, 20,34,15, 19, 20, 21, 24,19, 16, 22, 16 ,18, 20,16,19

We observe that the value 19 has the maximum frequency i.e. 5.    Therefore, mode of the given data is 19.

Question. The arithmetic mean of the following data is 25, find the value of k.

Solution.  

Given mean = 25 sum /N = 25 15k + 390 = 25k + 350 25k – 15k = 40 10k = 40 k = 4

Question. Find the value of p for the following distribution whose mean is 16.6.

Question. Calculate the median from the following data:

Question. Draw an ogive to represent the following frequency distribution:

Solution.  The given frequency distribution is not continuous, so we will first make it continuous and then prepare the cumulative frequency:

Plot the points (4.5, 2), (9.5, 8), (14.5, 18), (19.5, 23), (24.5,26) by taking the upper class limit over the x-axis and cumulative frequency over the y-axis.

Question. To find out the concentration of SO 2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

find the mean concentration of SO 2 in the air. Solution.  We may find class marks for each interval by using the relation

Class size of this data = 0.04 let a = 0.14

Question. Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0,1,2,3,4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss

Solution.  Let the assumed mean (A) = 2

Mean number of per toss = 2 + 470/1000 = 2 + 0.47 = 2.47

Question. If ∑fi = 11, ∑fixi = 2p + 52 and the mean of any distribution is 6, find the value of p.  Solution.

Question. Compare the modal ages of two groups of students appearing for an entrance test:

For Group A: Here the maximum frequency is 78, then the corresponding class 18 – 20 is modal class.

For group B: Here the maximum frequency is 89, then the corresponding class 18 – 20 is the modal class.

Hence the modal age for the Group A is higher than that for Group B.

Question. The following distribution gives the daily income of 50 workers of a factory :

Convert the distribution to a ‘less than type’ cumulative frequency distribution and draw its ogive. Hence obtain the median of daily income.

Hence, Median of daily income = Rs 345

Question. The mean of a set of numbers is . If each number is multiplied by k, then find the mean of the new set.  Solution.

Question. Find the value of x, if the mode of the following data is 25: 15, 20, 25,18,14,15, 25,15,18,16, 20, 25, 20, x, 18 Solution.  The frequency table of the given data is as given below:

It is given that the mode of the given date is 25. So, it must have the maximum frequency. That is possible only when x =25. Hence, x = 25.

Question. The annual profits earned by 30 shops of a shopping complex in a locality are recorded in the table shown below:

If we draw the frequency distribution table for the above data, find the frequency corresponding to the class 20-25. Solution.  The frequency table is as follows:

The frequency corresponding to the class 20 – 25 is 4. 12

Question. From the following information, construct less than and more than Ogive and find out median from it.

Question. A set of numbers consists of four 5’s, six 7’s, ten 9’s, eleven 12’s, three 13’s, two 14’s. Find the mode of this set of numbers. Solution.  Mode = 12

Since, It has highest frequency

Assignments for Class 10 Mathematics Statistics as per CBSE NCERT pattern

All students studying in Grade 10 Mathematics Statistics should download the assignments provided here and use them for their daily routine practice. This will help them to get better grades in Mathematics Statistics exam for standard 10. We have made sure that all topics given in your textbook for Mathematics Statistics which is suggested in Class 10 have been covered ad we have made assignments and test papers for all topics which your teacher has been teaching in your class. All chapter wise assignments have been made by our teachers after full research of each important topic in the textbooks so that you have enough questions and their solutions to help them practice so that they are able to get full practice and understanding of all important topics. Our teachers at https://www.assignmentsbag.com have made sure that all test papers have been designed as per CBSE, NCERT and KVS syllabus and examination pattern. These question banks have been recommended in various schools and have supported many students to practice and further enhance their scores in school and have also assisted them to appear in other school level tests and examinations. Its easy to take print of thee assignments as all are available in PDF format.

Some advantages of Free Assignments for Class 10 Mathematics Statistics

• Solving Assignments for Mathematics Statistics Class 10 helps to further enhance understanding of the topics given in your text book which will help you to get better marks
• By solving one assignments given in your class by Mathematics Statistics teacher for class 10 will help you to keep in touch with the topic thus reducing dependence on last minute studies
• You will be able to understand the type of questions which are expected in your Mathematics Statistics class test
• You will be able to revise all topics given in the ebook for Class 10 Mathematics Statistics as all questions have been provided in the question banks
• NCERT Class 10 Mathematics Statistics Workbooks will surely help you to make your concepts stronger and better than anyone else in your class.
• Parents will be able to take print out of the assignments and give to their child easily.

All free Printable practice assignments are in PDF single lick download format and have been prepared by Class 10 Mathematics Statistics teachers after full study of all topics which have been given in each chapter so that the students are able to take complete benefit from the worksheets. The Chapter wise question bank and revision assignments can be accessed free and anywhere. Go ahead and click on the links above to download free CBSE Class 10 Mathematics Statistics Assignments PDF.

Free PDF download of Statics Class 10 Maths Assignment Chapter wise pdf created by master educators from the latest syllabus of CBSE Boards. By practicing these Maths Assignment for Class 10 Chapter Wise PDF will help you to score more marks in your CBSE Board Examinations. We also give free NCERT Solutions and other study materials for students to make their preparation better.

Students who are searching for better solutions can download the Statics Class 10 Maths Assignment Chapter wise pdf with answers to assist you with revising the whole syllabus and score higher marks in your exam.

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You can get free PDF downloadable assignments for Grade 10 Mathematics Statistics from our website which has been developed by teachers after doing extensive research in each topic.

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NCERT Solutions for Class 10 Maths Chapter 14 Probability

NCERT Solutions for Maths Chapter 14 Probability Class 10 - Free PDF Download

Class 10 Maths NCERT Solutions for Chapter 14 helps students understand the concept of set theory, like unions and intersections, which can be applied to represent events and analyse their probabilities. Probability provides a way of quantifying uncertainty and helps in making informed decisions based on the likelihood of various events. Understanding and applying probability concepts is essential for solving real-world problems effectively.

The free PDF of Class 10 Maths Chapter 14 Solutions PDF download is available on Vedantu, giving students a better understanding of the problems. It covers solutions to every exercise in this chapter and is updated according to the latest CBSE syllabus . 

Glance on Maths Chapter 14 Class 10 - Probability

Chapter 14 of Class 10 Maths deals with the foundation by defining probability as the measure of how likely an event is to occur. It introduces the concept of outcomes, events, and sample spaces.

You'll learn how to identify favourable outcomes (those that satisfy a specific condition) and calculate the probability of an event by dividing the number of favourable outcomes by the total number of possible outcomes.

Probability in class 10 explains Fundamental theorems like the addition and multiplication rules of probability are introduced. These rules help you calculate the probability of combined events.

This article contains chapter notes, important questions, exemplar solutions, and exercise links for Chapter 14 - Probability, which you can download as PDFs.

There is one exercise (25 fully solved questions) in class 10th maths chapter 14 Probability.

Access Exercise wise NCERT Solutions for Chapter 14 Maths Class 10

Exercises under NCERT Solutions for Class 10 Maths Chapter 14 Probability

Exercise 14.1: This exercise involves solving word problems by paying close attention to key terms like event, sample space, and favourable outcome definitions. Break down complex problems into simpler events and apply the relevant rules.

Access NCERT Solutions for Class 10 Maths Chapter 14 – Probability

Exercise 14.1.

1. Complete the following statements:

i. Probability of an event E + Probability of the event ‘not E’ = _____. 

If the probability of an event be $p$, then the probability of the ‘not event’ will be, $1-p$ . Thus, the sum will be, $p+1-p=1$.

ii. The probability of an event that cannot happen is _____. Such an event is called _____.

The probability of an event that cannot happen is always $0$. iii. The probability of an event that is certain to happen is _____. Such an event is called _____.

The probability of an event that is certain to happen is $1$ . Such an event is called, sure event.

iv. The sum of the probabilities of all the elementary events of an experiment is _____.

The sum of the probabilities of all the elementary events of an experiment is $1$.

v. The probability of an event is greater than or equal to and less than or 

equal to _____.

The probability of an event is greater than or equal to $0$  and less than or equal to $1$ .

2. Which of the following experiments have equally likely outcomes? Explain.

i. A driver attempts to start a car. The car starts or does not start.

Equally likely outcomes defined as the outcome when each outcome is likely to occur as the others. So, the outcomes are not equally likely outcome.

ii. A player attempts to shoot a basketball. She/he shoots or misses the shot. 

The outcomes are not equally likely outcome.

(iii) A trial is made to answer a true-false question. The answer is right or wrong. 

The outcomes are equally likely outcome.

(iv) A baby is born. It is a boy or a girl. 

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

We already know the fact that a coin has only two sides, head and tail. So, when we toss a coin, it will either give us the result head or tail. There is no chance of the coin landing on his edge. And on the other hand, the chances of getting head and tail are also just the same. So, it can be concluded that the tossing of a coin is a fair way to decide the utcome, as it can not be biased and both teams will have the same chance of winning.

4. Which of the following cannot be the probability of an event?

(A) $\frac{2}{3}$

The probability of an event have to always be in the range of $[0,1]$ .

Now, let us the check the given values.

We can see, $\frac{2}{3}=0.67$ . This is in the given range. It can be a probability of an event.

We can see, $-1.5$ , which is a negative number and not inside the given range. It can not be a probability of an event.

We can see, $15%=\frac{15}{100}=0.15$ . This is in the given range of $[0,1]$ . It can be a probability of an event.

(D) $0.7$ 

We can see, $0.7$ , which is in the given range. It can be a probability of an event.

5. If $P(E)=0.05$ , what is the probability of an event ‘not $E$’? 

The sum of the probabilities of all events in always $1$ .

Thus, if $P(E)=0.05$ , the probability of the event ‘not E’ is, $1-0.05=0.95$.

6. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

i.an orange flavored candy?

There is no orange candy available in the bag, so, the probability of taking out an orange flavored candy is $0$.

(ii) a lemon flavored candy? 

All the candies in the bag are lemon flavored candies only. Thus, any candy Malini takes out will be a lemon flavored candy. 

So, the probability of taking out a lemon flavored candy is $1$.

7. It is given that in a group of 3 students, the probability of $2$  students not having the same birthday is $0.992$ . What is the probability that the $2$  students have the same birthday?

It is provided to us that, probability of 2 students not having the same birthday is, $0.992$ .

So,$P(2\text{ students }\text{having }\text{the}\text{ same}\text{ birthday})+P(\text{2}\text{ students }\text{not }\text{having }\text{the}\text{ same}\text{ birthday})=1$

$\Rightarrow P(\text{2}\text{ students}\text{ having}\text{ the }\text{same}\text{ birthday})+0.992=1$

Simplifying further,

$P(2 \text{students having the same birthday})=0.008$.

8. A bag contains $3$  red balls and $5$  black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ?

The bag is having $3$ red balls and $5$ black balls. 

Now, the probability of getting a red ball will be, $\frac{number\,of\,red\,balls}{total\,number\,of\,balls}$ .

Putting the values, we get, $\frac{3}{3+5}=\frac{3}{8}$ .

(ii) not red? 

And, the probability of getting a red ball will be, $1-\frac{number\,of\,red\,balls}{total\,number\,of\,balls}$.

Again, putting the values, $1-\frac{3}{8}=\frac{5}{8}$.

9. A box contains $5$ red marbles, $8$ white marbles and green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ?

The box is containing, $5$ red marbles, $8$ white marbles and green marbles.

The probability of getting a red marble will be, $\frac{number\,of\,red\,marbles}{total\,number\,of\,marbles}$

Putting the values, $\frac{5}{5+8+4}=\frac{5}{17}$.

(ii) white ? 

Again, the probability of getting a white marble will be, $\frac{number\,of\,white\,marbles}{total\,number\,of\,marbles}$

Putting the values, $\frac{8}{5+8+4}=\frac{8}{17}$.

(iii) not green? 

And, the probability of getting a green marble will be, $\frac{number\,of\,green\,marbles}{total\,number\,of\,marbles}$.

Putting the values, $\frac{4}{5+8+4}=\frac{4}{17}$ .

So, the probability of the marble not being green will be, $1-\frac{4}{17}=\frac{13}{17}$.

10. A piggy bank contains hundred $50$ p coins, fifty ` $1$  coins, twenty ` $2$  coins and ten ` $5$  coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a $50$  p coin ?

We are provided with the fact that, the piggy bank contains, hundred $50$ p coins, fifty $1$ rs coins, twenty $2$ rs coins and ten $5$ rs coins.

So, the total number of coins, $100+50+20+10=180$ .

Thus, the probability of drawing a $50$ p coin, $\frac{100}{180}=\frac{5}{9}$.

ii. will not be a ` $5$  coin?

Similarly, the probability of drawing a $5$ rs coin, $\frac{10}{180}=\frac{1}{18}$ .

Thus, the probability of not getting a $5$ rs coin, $1-\frac{1}{18}=\frac{17}{18}$ .

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing $5$  male fish and $8$  female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?

The total number of fishes in the tank, $5+8=13$ .

Thus, the probability of getting a male fish, $\frac{no\,of\,male\,fishes}{total\,no\,of\,fishes}=\frac{5}{13}$.

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1,2,3,4,5,6,7,8$  (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at

We can see, there are $8$  numbers on spinner then the total number of favorable outcome is $8$ .

Thus, the probability of getting the number $8$ is, $\frac{no\,of\,digit\,8\,on\,the\,spinner}{total\,number\,of\,digits}=\frac{1}{8}$.

(ii) $\text{an odd number?}$

There are 4 odd digits, $1,3,5,7$ .

Thus, the probability of getting an odd number is,  

$\frac{no\,of\,odd\,digits\,}{total\,number\,of\,digits}=\frac{4}{8}=\frac{1}{2}$ .

(iii) $\text{a number greater than 2?}$ 

There are 6 numbers greater than 2, say $3,4,5,6,7,8$ .

Thus, the probability of getting a number greater than 2, $\frac{no\,of\,digits\,greater\,than\,2\,on\,the\,spinner}{total\,number\,of\,digits}=\frac{6}{8}=\frac{3}{4}$.

(iv) $\text{a number less than 9}$$?$ 

As we can see, every number in the spinner is less than $9$ , thus, we get, 

The probability of getting a number less than $9$ , will be, $1$.

13. A die is thrown once. Find the probability of getting (i) a prime number; 

There is $6$ results can be obtained from a dice. 

There are 3 prime numbers, $2,3,5$ among those results.

Thus, the probability of getting a prime number,$=\frac{\text{no }\text{of }\text{prime }\text{numbers }\text{in }\text{a}\text{ dice}}{\text{total}\text{ numbers}\text{ on }\text{dice}}=\frac{3}{6}=\frac{1}{2}$

(ii) a number lying between $2$ and $6$ 

There are 3 numbers between 2 and 6, 3,4,5.

Thus, the probability of getting a number between 2 and 6,

$=\frac{\text{no }\text{of }\text{numbers }\text{between }\text{2}\text{ and}\text{ 6 }\text{in}\text{ a}\text{ dice}}{\text{total}\text{ numbers }\text{on}\text{ dice}}=\frac{3}{6}=\frac{1}{2}$

(iii) an odd number. 

There are 3 odd numbers among the results, 1,3,5.

Thus, the probability of getting a odd number,

$=\frac{\text{no }\text{of}\text{ odd }\text{numbers}\text{ between}\text{ in }\text{a }\text{dice}}{\text{total}\text{ numbers }\text{on}\text{ dice}}=\frac{3}{6}=\frac{1}{2}$

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red color

We know there are 52 numbers in the deck.

There are 2 kings of red color in the deck.

Thus, the probability,

$=\frac{total\,number\,of\,red\,kings}{total\,number\,of\,cards}=\frac{2}{52}=\frac{1}{26}$

(ii) a face cards 

There are 12 face cards in the deck.

$=\frac{total\,number\,of\,face\,cards}{total\,number\,of\,cards}=\frac{12}{52}=\frac{3}{13}$

(iii) a red face cards 

There are 6 red face cards in the deck.

$=\frac{total\,number\,of\,red\,face\,cards}{total\,number\,of\,cards}=\frac{6}{52}=\frac{3}{26}$

(iv) the jack of hearts

There are 1 jack of hearts card in the deck.

$=\frac{total\,number\,of\,red\,face\,cards}{total\,number\,of\,cards}=\frac{1}{52}$

(v) a spade

There are 13 spade cards in the deck.

$=\frac{total\,number\,of\,spade\,cards}{total\,number\,of\,cards}=\frac{13}{52}=\frac{1}{4}$

(vi) the queen of diamonds 

There are 1 queen of diamonds card in the deck.

15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

There are total 5 cards given in our deck.

Thus, the probability of getting a queen card among the 5 cards,

\[=\frac{number\,of\,queen}{number\,of\,total\,cards}=\frac{1}{5}\]

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen? 

Now, the queen is put aside, so there will be 4 cards left.

a. Thus, the probability of getting a queen card among the 5 cards,

\[=\frac{number\,of\,ace}{number\,of\,total\,cards}=\frac{1}{4}\] 

b. There are no queen cards left in the deck.

Thus, the probability of getting a queen card among the 4 cards,

\[=\frac{number\,of\,queen}{number\,of\,total\,cards}=\frac{0}{4}=0\]

16. 12 defective pens are accidentally mixed with 132  good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

There are total (132+12)=144 number of pens in the lot.

And also there are 132 good pens in the given collection.

Thus, the probability of getting a good pen,

$=\frac{number\,of\,good\,pens}{number\,of\,total\,pens}=\frac{132}{144}=\frac{11}{12}$

17. A lot of 20  bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

There are 4 defective bulbs among 20 bulbs.

Thus, the probability of getting a defective bulb,

$=\frac{number\,of\,defective\,bulb}{number\,of\,total\,bulb}=\frac{4}{20}=\frac{1}{5}$

ii. Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?

After the first draw, there are 19 bulbs left in the lot. Again, as the bulb was a non-defective bulb, the total non-defective bulbs are 15.

Therefore, the probability of not getting a defective bulb this time,

$=\frac{15}{19}$ .

18. A box contains 90 discs which are numbered from $1$  to $90$ . If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number. 

There are 81 two digit numbers between 1 to 90.

Thus, the probability of getting a two digit number in the draw,

$=\frac{the\,total\,number\,of\,two\,digit\,numbers}{total\,numbers}=\frac{81}{90}=\frac{9}{10}$

(ii) a perfect square number 

The number of perfect square numbers between 1 to 90.

Thus, the probability of getting a perfect number in the draw,

$=\frac{the\,total\,number\,of\,perfect\,number}{total\,numbers}=\frac{9}{90}=\frac{1}{10}$

(iii) a number divisible by $5$. 

The number of numbers divisible by 5, 18.

Thus, the probability of getting a number divisible by 5 in the draw,

$=\frac{the\,total\,number\,of\,number\,divisible\,by\,5}{total\,numbers}=\frac{18}{90}=\frac{1}{5}$

19. A child has a die whose six faces show the letters as given below: A, A, B, C, D, E. The die is thrown once. What is the probability of getting

(i) A? 

There are two A’s in the six faces, so, the probability of getting an A,

$=\frac{total\,number\,of\,A's}{total\,number\,of\,sides}=\frac{2}{6}=\frac{1}{3}$

(ii)  D? 

There are one D in the six faces, so, the probability of getting an D,

$=\dfrac{total\,number\,of\,D's}{total\,number\,of\,sides}=\dfrac{1}{6}$

20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?

It is given that it is a rectangle with sides 3 m and 2 m.

Thus, the area of the rectangle,

$=$ length $\times$ breadth

$=3 \times 2=6 \mathrm{~m}^{2}$

The radius of the circle, half of diameter $=\frac{1}{2}\,m$ .

The area of the circle, $=\pi .{{\left( \frac{1}{2} \right)}^{2}}=\frac{\pi }{4}\,{{m}^{2}}$

Thus, the probability of the die landing inside the circle is,

$=\frac{area\,of\,the\,circle}{area\,of\,the\,rectangle}=\frac{\frac{\pi }{4}}{6}=\frac{\pi }{24}$

21. A lot consists of $144$  ball pens of which $20$ are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it ? 

There are total 144 ball pens in the lot and 20 of them are defective.

Thus, the total number of non-defective pens, $(144-20)=124$ .

Nuri will not buy the pen if it is defective, thus,

The probability of getting a good pen is,

$=\frac{total\,no\,of\,good\,pens}{total\,no\,of\,pens}=\frac{124}{144}=\frac{31}{36}$

(ii) She will not buy it ? 

Now, the probability of Nuri not buying the pen is,

$=1-\frac{31}{36}=\frac{36-31}{36}=\frac{5}{36}$

22. Refer to example 13: (i) Complete the following table:

If there are two dices thrown simultaneously, then we can get the following results,

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Thus, the total number of results is 36.

Probability of getting a sum of 2, $=\dfrac{1}{36}$ 

Probability of getting a sum of 3, $=\dfrac{2}{36}=\dfrac{1}{18}$ 

Probability of getting a sum of 4, $=\dfrac{3}{36}=\dfrac{1}{12}$ 

Probability of getting a sum of 5, $=\dfrac{4}{36}=\dfrac{1}{9}$ 

Probability of getting a sum of 6, $=\dfrac{5}{36}$ 

Probability of getting a sum of 7, $=\dfrac{6}{36}=\dfrac{1}{6}$ 

Probability of getting a sum of 8, $=\dfrac{5}{36}$ 

Probability of getting a sum of 9, $=\dfrac{4}{36}=\dfrac{1}{9}$ 

Probability of getting a sum of 10, \[=\dfrac{3}{36}=\dfrac{1}{12}\] 

Probability of getting a sum of 11, $=\dfrac{2}{36}=\dfrac{1}{18}$ 

Probability of getting a sum of 12, $=\dfrac{1}{36}$ 

Thus, we get the values of our table.

(ii) A student argues that there are $11$ possible outcomes$2,3,4,5,6,7,8,9,10,11\text{ and 12}$ .Therefore, each of them has a probability $\frac{1}{11}$ . Do you agree with this argument?

Justify your answer.

As we can see different values all over the table, we can conclude that, the given statement is wrong. The probability of each of them can never be $\dfrac{1}{11}$. 23. A game consists of tossing a one rupee coin $3$  times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Hanif will win if he gets 3 heads and 3 tails consecutively.

The probability of Hanif losing the game, = The probability of not getting 3 heads and 3 tails.

The possible outcomes of the tosses, $(HHH,HHT,HTH,HTT,THH,THT,TTH,TTT)$ 

The total number of outcomes is, 8.

Thus, the probability of not getting 3 heads and 3 tails, 

$=1-\frac{2}{8}=\dfrac{6}{8}=\dfrac{3}{4}$.

24. A die is thrown twice. What is the probability that

(i) $5$  will not come up either time? 

(Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment)

We can see that two dices are thrown altogether, thus the total number of outcomes =36.

Now, the total cases where atleast 5 occurs is, 11, i.e,$(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)$  

So, the probability of not getting 5 either time is,$=1-\frac{11}{36}=\frac{25}{36}$.

(ii) $5$  will come up at least once?

And, probability of getting 5 atleast once is,$\frac{11}{36}$.

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $\frac{1}{3}$. 

In this problem, two coins are tossed simultaneously, thus we get 4 outcomes, i.e, $(HH,HT,TH,TT)$ .

So, the probability of getting both heads and tails, $=\frac{2}{4}=\frac{1}{2}$ .

Thus, the statement is wrong.

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is $\frac{1}{2}$. 

By throwing a die, we get 6 possible outcomes.

The odd numbers are $1,3,5$ .

Thus the probability of getting a odd number, $=\frac{3}{6}=\frac{1}{2}$ .

So, the statement is true.

Overview of Deleted Syllabus for CBSE Class 10 Maths Probability

Vedantu's NCERT Solutions for Class 10 Maths Chapter 14 - Probability offers an exceptional resource for students seeking to grasp the complexities of probability theory. 

You will be able to solve problems involving coin tosses, dice rolls, card games, and other scenarios where chance plays a role.

With comprehensive and well-structured explanations, the platform empowers learners to tackle real-world challenges with confidence. 

Other Study Material for CBSE Class 10 Maths Chapter 14

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 10 Maths Chapter 14 Probability

1. What is the curriculum of CBSE Board Class 10 Mathematics?

There are 15 units in Class 10 Mathematics. These include Arithmetic Progressions, Pair of Linear Equations in Two Variables, Real numbers, Polynomials, Quadratic Equations, Triangles, Some Applications of Trigonometry, Introduction to Trigonometry, Constructions, Circles, Coordinate Geometry, Surface Areas and Volumes, Areas Related to Circles, Probability and Statistics.

2. What are the applications of probability in day to day life?

Probability is a branch of mathematics that deals with the likelihood of events happening. It is used in a wide variety of applications in our daily lives, including:

Weather forecasting: Weather forecasters use probability to estimate the likelihood of rain, snow, or other types of precipitation.

Sports betting: Sports bettors use probability to assess the likelihood of a particular team or player winning a game.

Insurance: Insurance companies use probability to calculate the risk of an event happening and to determine the premiums that they charge their customers.

Medical diagnosis: Doctors use probability to assess the likelihood of a patient having a particular condition.

Stock market investing: Stock market investors use probability to assess the risk of an investment and to determine the potential returns.

Marketing: Marketers use probability to target their advertising campaigns to the most likely customers.

Decision making: In many situations, we need to make decisions based on uncertain information. Probability can help us to make more informed decisions by providing us with a framework for assessing the likelihood of different outcomes.

Here are some other examples of how probability is used in our daily lives:

When we decide what to wear, we are essentially using probability to assess the likelihood of different weather conditions.

When we choose a route to drive to work, we are using probability to assess the likelihood of traffic congestion.

When we decide what to eat, we are using probability to assess the likelihood of different foods being available at the grocery store.

When we decide what to watch on TV, we are using probability to assess the likelihood of different shows being interesting to us.

3. What are the subtopics of Ch 14 Maths Class 10?

Subtopics of Chapter 14 of the Class 10 Maths NCERT textbook:

Introduction

Experiment and Outcome

Sample Space

Elementary Events

Favourable Outcomes

Probability of an Event

Theoretical Probability

Experimental Probability

Addition Theorem of Probability

Complement of an Event

Multiplication Theorem of Probability

This chapter introduces the concept of probability, which is the likelihood of an event occurring. It discusses the different types of events, such as elementary events, favourable outcomes, and sure events. The chapter also covers the two methods of calculating probability: theoretical probability and experimental probability. The addition and multiplication theorems of probability are also discussed in this chapter.

4. What can we learn in the chapter Probability?

In the chapter Probability, we can learn about the following:

The definition of probability and how it is measured.

The different types of events, such as elementary events, favourable outcomes, and sure events.

The two methods of calculating probability: theoretical probability and experimental probability.

The addition and multiplication theorems of probability.

How to use probability to make predictions about the outcomes of random events.

Here are some specific things we can learn:

The probability of an event can be anywhere from 0 to 1. A probability of 0 means that the event is impossible, while a probability of 1 means that the event is certain.

The probability of an event can be calculated by dividing the number of favourable outcomes by the total number of possible outcomes.

If we repeat an experiment many times, the experimental probability of an event will approach the theoretical probability of the event.

The addition theorem of probability can be used to calculate the probability of two events occurring when the events are mutually exclusive.

The multiplication theorem of probability can be used to calculate the probability of two events occurring when the events are not mutually exclusive.

5. State the basic law behind probability?

The law of probability informs us how likely particular occurrences are to occur. According to the rule of large numbers, the more trials you have in an experiment, the closer you come to a precise probability. The multiplication rule is used to calculate the likelihood of two occurrences occurring at the same time. 

6. State an event where the probability is ½?

On tossing a random fair coin, there are two possible outcomes i.e. head and tail. 

When a coin is tossed, the probability of getting ahead is ½ and the probability of getting a tail is also ½. This is based on the probability of the specific event occurring.

7. What is the chapter probability about?

Probability is a field of mathematics that deals with numerical descriptions of what will happen in the future, or whether something is true or not. This is a high-scoring yet challenging Chapter in Class 10 Maths. As a result, you must be conversant with the tips and methods required to swiftly answer numerical issues. In this regard,  Vedantu  offers exact NCERT Solutions for Class 10 Maths Probability, which include all types of sums that may be encountered in examinations.

8. Why choose Vedantu for Chapter 14 Class 10 Maths?

In a highly competitive world, students are thriving for the best educational services to score as much as they can, Vedantu offers the best solutions. NCERT Solutions for Class 10 Maths Probability is one of the best study guides for students, to help them achieve their desired scores. These solution PDFs are available at free of cost on the Vedantu app and the Vedantu website.

9. What is the main focus of Chapter 14 - Probability Class 10 Maths?

The main focus of Chapter 14 - Probability Class 10 is to introduce students to the basic concepts of probability, including the theoretical approach, experiments, outcomes, events, and the probability of an event occurring.

10. How does NCERT Solutions for Class 10 Probability Chapter 14 help in understanding probability?

The NCERT Solutions for class 10 probability provides detailed explanations and step-by-step solutions to textbook problems, helping students understand the fundamental principles of probability and how to apply them to solve various problems.

11. What key concepts are covered in the NCERT Solutions for Probability Class 10 solutions Chapter 14?

Key concepts covered in class 10 probability include:

Probability of an event

Complementary events

Probability of simple and compound events

Use of probability in real-life situations

12. Are the probability class 10 solutions aligned with the latest syllabus?

Yes, the NCERT probability class 10 solutions are aligned with the latest CBSE syllabus for the academic year 2024-25, ensuring that all topics and concepts are covered as per the curriculum.

13. How do probability class 10 solutions assist in exam preparation?

The probability class 10 solutions include a variety of problems and their detailed solutions, which help students practice extensively. This thorough practice enhances problem-solving skills and prepares students effectively for exams.

14. Can these solutions help in clearing doubts about class 10 probability concepts?

Absolutely. In probability class 10 NCERT solutions, detailed and clear explanations provided in the solutions help in resolving any doubts or misconceptions students might have about probability concepts.

15. What types of questions are included in the probability class 10 NCERT solutions?

The probability class 10 NCERT solutions include a wide range of questions such as multiple-choice questions, short answer questions, and long answer questions. These questions cover theoretical concepts, practical problems, and real-life applications of probability.

16. How can students benefit from practising probability class 10 NCERT solutions regularly?

Regular practice of chapter 14 class 10 maths helps students build a strong foundation in probability, enhances their problem-solving abilities, and boosts their confidence in tackling various types of questions in exams.

17. Are there any real-life examples provided in the solutions to explain chapter 14 class 10 maths concepts?

Yes, the chapter 14 class 10 maths solutions include real-life examples to illustrate how probability is used in everyday situations, making the concepts more relatable and easier to understand.

18. Where can students access the class 10 maths ch 14 for Class 10 Maths Chapter 14 - Probability?

Students can access the class 10 maths ch 14 solutions on educational platforms like Vedantu, where they are available for free download in PDF format, providing easy and convenient access for study and revision.

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Case Study on Statistics Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Statistics Class 10 Maths can use this page to download the PDF file. 

The case study questions on Statistics are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Statistics case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Statistics Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Statistics, therefore, they prepared a set of solutions along with the case study questions on Statistics.

The case study on Statistics Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Statistics case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Statistics Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Statistics case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Statistics Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Statistics case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Statistics Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Statistics as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Statistics?

Students can choose their own way to answer Case Study on Statistics Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Statistics Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Statistics questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Statistics Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Statistics?

 A few essential things to know to solve Case Study Questions on Class 10 Statistics are -

• Basic Formulas of Statistics: One of the most important things to know to solve Case Study Questions on Class 10 Statistics is to learn about the basic formulas or revise them before solving the case-based questions on Statistics.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Statistics case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Statistics Class 10 Maths?

Use Selfstudys.com to find Case Study on Statistics Class 10 Maths. For ease, here is a step-wise procedure to download the Statistics Case Study for Class 10 Maths in PDF for free of cost.

Since you are already on this page, you can scroll to the top section of this page to get access to the Case Study on Statistics. To help others reach this page let them know these steps:

• Open Selfstudys.com on your computer/laptop or Smartphone 
• Once the website gets loaded, click on the navigation button

• Find CBSE from the given menu

• Click on Case Study

• Choose Class 10 
• Search Maths and then navigate to the Statistics Class 10 Maths Case Study

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Linear programming class xii chapter 12, math assignment ch-13 class x | statistics.

Chapter: 13(Statistics) Formulas used to solve the Assignment on statistics

Range =  Highest value - Lowest value

Class size =  Upper limit - Lower limit

MODE of Data in Statistics

W here l = lower limit of the modal class,

 h = size of the class interval 

  f 1  = frequency of the modal class

f 0  = frequency of the class preceding the modal class,

f 2  = frequency of the class succeeding the modal class.

Median of Data in Statistics

Where l= lower limit of median class                           

N = Sum of all observations

f = frequency of the median class.       

Cf= cumulative frequency of class  preceding the median class

 h= height of  median class

Empirical Formula of Data in Statistics

      Mode= 3 Median  - 2 Mean

Empirical formula does not provide the same result in different situations. So, students should use it only when asked otherwise avoid it.

* If median class is the first class –interval then cumulative frequency (Cf) of the preceding class interval should be taken as zero.

* For finding Mean class interval need not to be continuous.

* For finding Mode and Median class interval should be continuous. 

* For making class interval continuous we should subtract 0.5 from the lower limits and add 0.5 to the upper limits of all class intervals. 

Assignment on Statistics Solve the following questions

Problems based on mean.

Do we use here step-deviation method ? Why or Why not   

No, we cannot use step deviation method here because height of the class intervals are not equal. So we can solve this question either by Direct Method or by Assumed mean method

PROBLEMS BASED ON MODE  

QUESTION 10. For the given data find mean & mode.              

Ans : Mean=26 & Mode=22.85

PROBLEMS BASED ON MEDIAN

QUESTION 15. Find median.         Ans: 40

Ans:  f 1 =35 &  f 2 = 25

MISCELLANEOUS PROBLEMS ON MEAN, MODE, MEDIAN  

QUESTION  21. 

If mode = 24.5 and mean is 29.75 then find median by using empirical formula.    [Ans: 28]

QUESTION 23. Calculate Mean, mode, median.     Ans : Mean = 51.63, Mode = 53.33, Median = 52

Marks no of students more than 0 80 more than 10 76 more than 20 71 more than 30 65 more than 40 57 more than 50 43 more than 60 28 more than 70 15 more than 80 10 more than 90 8 more than 100 0, higher order thinking skills (hots).

Answer:  f 1  = 28,  f 2  = 32 and  f 3  = 24

Solution Hint:   Use direct method to solve this question

For easy calculation use step deviation method and take assume mean (a) in front of f 1

QUESTIONS DELETED FROM CBSE SYLLABUS

QUESTION 25. Find median by using less than ogive & actual calculation.                               

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