NCERT Solutions for Class 11 Maths Chapter 15 Statistics

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.

The NCERT Solutions for Class 11 Maths Chapter 15 Statistics are available in PDF format on this page. The NCERT Solutions are crafted by the most experienced educators in the teaching industry, according to the latest CBSE Syllabus 2023-24, making the solution to every problem straightforward and justifiable. Every solution, written in the PDF given below, helps students to get ready for the exam and score good marks. These solutions assist a Class 11 student in mastering the concepts of statistics that are categorised under Class 11 Maths.

NCERT Solutions for Class 11 Maths Chapter 15 Statistics

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Get detailed solutions for all the questions from the below links.

Exercise 15.1 Solutions: 12 Questions

Exercise 15.2 Solutions: 10 Questions

Exercise 15.3 Solutions: 5 Questions

Miscellaneous Exercise Solutions: 7 Questions

Access Solutions for Class 11 Maths Chapter 15 Statistics

Exercise 15.1 Page: 360

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

Solution:-

First, we have to find (x̅) of the given data.

So, the respective values of the deviations from mean,

i.e., x_i – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 9 = 1, 10 – 10 = 0,

10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7

6, 3, 2, 1, 0, -2, -3, -7

Now, absolute values of the deviations,

6, 3, 2, 1, 0, 2, 3, 7

MD = sum of deviations/ number of observations

= 24/8

= 3

So, the mean deviation for the given data is 3.

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution:-

First, we have to find (x̅) of the given data.

So, the respective values of the deviations from mean,

i.e., x_i – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,

50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6

-12, 20, -2, -10, -8, 5, 13, -4, 4, -6

Now, absolute values of the deviations,

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

MD = sum of deviations/ number of observations

= 84/10

= 8.4

So, the mean deviation for the given data is 8.4.

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:-

First, we have to arrange the given observations into ascending order.

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

The number of observations is 12.

Then,

Median = ((12/2)^th observation + ((12/2)+ 1)^th observation)/2

(12/2)^th observation = 6^th = 13

(12/2)+ 1)^th observation = 6 + 1

= 7^th = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

Mean Deviation

= (1/12) × 28

= 2.33

So, the mean deviation about the median for the given data is 2.33.

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution:-

First, we have to arrange the given observations into ascending order.

36, 42, 45, 46, 46, 49, 51, 53, 60, 72.

The number of observations is 10.

Then,

Median = ((10/2)^th observation + ((10/2)+ 1)^th observation)/2

(10/2)^th observation = 5^th = 46

(10/2)+ 1)^th observation = 5 + 1

= 6^th = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Mean Deviation

= (1/10) × 70

= 7

So, the mean deviation about the median for the given data is 7.

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5.

Solution:-

Let us make the table of the given data and append other columns after calculations.

The sum of calculated data,

The absolute values of the deviations from the mean, i.e., |x_i – x̅|, as shown in the table.

6.

Solution:-

Let us make the table of the given data and append other columns after calculations.

Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

Solution:-

Let us make the table of the given data and append other columns after calculations.

Now, N = 26, which is even.

Median is the mean of the 13^th and 14^th observations. Both of these observations lie in the cumulative frequency of 14, for which the corresponding observation is 7.

Then,

Median = (13^th observation + 14^th observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are shown in the table.

8.

Solution:-

Let us make the table of the given data and append other columns after calculations.

Now, N = 29, which is odd.

So, 29/2 = 14.5

The cumulative frequency greater than 14.5 is 21, for which the corresponding observation is 30.

Then,

Median = (15^th observation + 16^th observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |x_i – M| are shown in the table.

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

Solution:-

Let us make the table of the given data and append other columns after calculations.

10.

Solution:-

Let us make the table of the given data and append other columns after calculations.

11. Find the mean deviation about median for the following data.

Solution:-

Let us make the table of the given data and append other columns after calculations.

The class interval containing N^th/2 or 25^th item is 20-30.

So, 20-30 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 20, c = 14, f = 14, h = 10 and n = 50

Median = 20 + (((25 – 14))/14) × 10

= 20 + 7.85

= 27.85

12. Calculate the mean deviation about median age for the age distribution of 100 persons given below.

[Hint: Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

Solution:-

The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.

The class interval containing N^th/2 or 50^th item is 35.5 – 40.5

So, 35.5 – 40.5 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100

Median = 35.5 + (((50 – 37))/26) × 5

= 35.5 + 2.5

= 38

Exercise 15.2 Page: 371

Find the mean and variance for each of the data in Exercise 1 to 5.

1. 6, 7, 10, 12, 13, 4, 8, 12

Solution:-

So, x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

Let us make the table of the given data and append other columns after calculations.

We know that the Variance,

σ² = (1/8) × 74

= 9.2

∴Mean = 9 and Variance = 9.25

2. First n natural numbers

Solution:-

We know that Mean = Sum of all observations/Number of observations

∴Mean, x̅ = ((n(n + 1))2)/n

= (n + 1)/2

and also, WKT Variance,

By substituting the value of x̅, we get

WKT, (a + b)(a – b) = a² – b²

σ² = (n² – 1)/12

∴Mean = (n + 1)/2 and Variance = (n² – 1)/12

3. First 10 multiples of 3

Solution:-

First, we have to write the first 10 multiples of 3.

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

So, x̅ = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

Let us make the table of the data and append other columns after calculations.

Then, the Variance

= (1/10) × 742.5

= 74.25

∴Mean = 16.5 and Variance = 74.25

4.

Solution:-

Let us make the table of the given data and append other columns after calculations.

5.

Solution:-

Let us make the table of the given data and append other columns after calculations.

6. Find the mean and standard deviation using short-cut method.

Solution:-

Let the assumed mean A = 64. Here, h = 1

We obtain the following table from the given data.

Mean,

Where A = 64, h = 1

So, x̅ = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, the variance,

σ² = (1²/100²) [100(286) – 0²]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

∴ Mean = 64 and Standard Deviation = 1.691

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

7.

Solution:-

Let us make the table of the given data and append other columns after calculations.

8.

Solution:-

Let us make the table of the given data and append other columns after calculations.

9. Find the mean, variance and standard deviation using the short-cut method.

Solution:-

Let the assumed mean, A = 92.5 and h = 5

Let us make the table of the given data and append other columns after calculations.

Mean,

Where, A = 92.5, h = 5

So, x̅ = 92.5 + ((6/60) × 5)

= 92.5 + ½

= 92.5 + 0.5

= 93

Then, the Variance,

σ² = (5²/60²) [60(254) – 6²]

= (1/144) [15240 – 36]

= 15204/144

= 1267/12

= 105.583

Hence, standard deviation = σ = √105.583

= 10.275

∴ Mean = 93, variance = 105.583 and Standard Deviation = 10.275

10. The diameters of circles (in mm) drawn in a design are given below.

Calculate the standard deviation and mean diameter of the circles.

[Hint: First, make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]

Solution:-

Let the assumed mean, A = 42.5 and h = 4

Let us make the table of the given data and append other columns after calculations.

Mean,

Where, A = 42.5, h = 4

So, x̅ = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, the Variance,

σ² = (4²/100²)[100(199) – 25²]

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553

∴ Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.

Exercise 15.3 Page: 375

1. From the data given below state which group is more variable, A or B?

Solution:-

For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

Co-efficient of variation (C.V.) = (σ/ x̅) × 100

Where, σ = standard deviation, x̅ = mean

For Group A.

Where A = 45,

and y_i = (x_i – A)/h

Here h = class size = 20 – 10

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

σ² = (10²/150²) [150(342) – (-6)²]

= (100/22500) [51,300 – 36]

= (100/22500) × 51264

= 227.84

Hence, standard deviation = σ = √227.84

= 15.09

∴ C.V for group A = (σ/ x̅) × 100

= (15.09/44.6) × 100

= 33.83

Now, for group B.

Where A = 45,

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

σ² = (10²/150²) [150(366) – (-6)²]

= (100/22500) [54,900 – 36]

= (100/22500) × 54,864

= 243.84

Hence, standard deviation = σ = √243.84

= 15.61

∴ C.V for group B = (σ/ x̅) × 100

= (15.61/44.6) × 100

= 35

By comparing the C.V. of group A and group B.

C.V of Group B > C.V. of Group A

So, Group B is more variable.

2. From the prices of shares X and Y below, find out which is more stable in value.

Solution:-

From the given data,

Let us make the table of the given data and append other columns after calculations.

We have to calculate Mean for x,

Mean x̅ = ∑x_i/n

Where, n = number of terms

= 510/10

= 51

Then, Variance for x =

= (1/10²)[(10 × 26360) – 510²]

= (1/100) (263600 – 260100)

= 3500/100

= 35

WKT Standard deviation = √variance

= √35

= 5.91

So, co-efficient of variation = (σ/ x̅) × 100

= (5.91/51) × 100

= 11.58

Now, we have to calculate Mean for y,

Mean ȳ = ∑y_i/n

Where, n = number of terms

= 1050/10

= 105

Then, Variance for y =

= (1/10²)[(10 × 110290) – 1050²]

= (1/100) (1102900 – 1102500)

= 400/100

= 4

WKT Standard deviation = √variance

= √4

= 2

So, co-efficient of variation = (σ/ x̅) × 100

= (2/105) × 100

= 1.904

By comparing C.V. of X and Y,

C.V of X > C.V. of Y

So, Y is more stable than X.

3. An analysis of monthly wages paid to workers in two firms, A and B, belonging to the same industry, gives the following results:

(i) Which firm, A or B, pays a larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

Solution:-

(i) From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 × 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 × 5253

= Rs 34,03,944

So, firm B pays larger amount as monthly wages.

(ii) Variance of firm A = 100

We know that, standard deviation (σ)= √100

=10

Variance of firm B = 121

Then,

Standard deviation (σ)=√(121 )

=11

Hence, the standard deviation is more in case of Firm B. That means, in firm B, there is greater variability in individual wages.

4. The following is the record of goals scored by team A in a football session:

For team B, the mean number of goals scored per match was 2, with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Solution:-

From the given data,

Let us make the table of the given data and append other columns after calculations.

C.V. of firm B is greater.

∴ Team A is more consistent.

5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

Which is more varying, the length or weight?

Solution:-

First, we have to calculate Mean for Length x.

Miscellaneous Exercise Page: 380

1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Solution:-

2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

Solution:-

3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Solution:-

We know that,

4. Given that x̅ is the mean and σ² is the variance of n observations x₁, x₂, …,x_n . Prove that the mean and variance of the observations ax₁, ax₂, ax₃, …., ax_n are ax̅ and a²σ², respectively, (a ≠ 0).

Solution:-

From the question, it is given that, n observations are x₁, x₂,…..x_n

Mean of the n observation = x̅

Variance of the n observation = σ²

As we know,

5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12

Solution:-

(i) If the wrong item is omitted,

From the question, it is given that

The number of observations, i.e., n = 20

The incorrect mean = 20

The incorrect standard deviation = 2

(ii) If it is replaced by 12,

From the question, it is given that

The number of incorrect sum observations, i.e., n = 200

The correct sum of observations n = 200 – 8 + 12

n = 204

Then, correct mean = correct sum/20

= 204/20

= 10.2

6. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry, are given below.

Which of the three subjects shows the highest variability in marks, and which shows the lowest?

Solution:-

From the question, it is given that

Mean of Mathematics = 42

Standard deviation of Mathematics = 12

Mean of Physics = 32

Standard deviation of physics = 15

Mean of Chemistry = 40.9

Standard deviation of chemistry = 20

As we know,

7. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on, it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Solution:-

From the question, it is given that

The total number of observations (n) = 100

Incorrect mean, (x̅) = 20

And, Incorrect standard deviation (σ) = 3

NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

The topics of Class 11 Chapter 15 – Statistics of NCERT Solution are as follows:

15.1 Introduction

This section talks about the concept of central tendency, mean and median [during even and the odd number of observations] with examples. It introduces the concept of measure of dispersion.

The values which cluster around the middle or centre of the distribution are measures of central tendency. They are mean, median and mode.

In a class, the mean can be used to find the average marks scored by the students.

When calculating the height of students, the median can be used to find the middlemost value.

15.2 Measures of Dispersion

This section defines measures of dispersion, different measures of dispersion [range, quartile deviation, mean deviation, standard deviation]

Measures of dispersion explain the relationship with measures of central tendency. For example, the spread of data tells how well the mean represents the data. If the spread is large, then the mean is not representative of the data.

15.3 Range

This section defines the range, its formula and an example.

The range gives the variability of scores using the maximum and minimum values in the set.

In a cricket match, Batsman A range = 121 – 0 = 121 and Batsman B range = 52 – 22 = 30

Range A > Range B. So, the scores are spread in the case of A, whereas for B, they are close to each other.

15.4 Mean Deviation

This section defines mean deviation, the formula for mean deviation.

The concept of mean deviation can be used by biologists in the comparison of different animal weights and deciding what would be a healthy weight.

15.4.1 Mean deviation for ungrouped data

The process of obtaining the mean deviation for ungrouped data is elaborated in this section.

Find the mean, deviations from the mean, and absolute deviations and substitute the values in the mean deviation formula and arrive at the answer.

15.4.2 Mean deviation for grouped data

The process of obtaining mean deviation for discrete and continuous frequency distributions is elaborated in this section with solved examples.

15.4.3 Limitations of mean deviation

• If, in a series, the degree of variability is very high, then the median will not be representative of the data. Hence, the mean deviation calculated about such a median cannot be relied on.
• If the sum of deviations from the mean is greater than the sum of deviations from the median, then the mean deviation from the mean is not very specific.
• The absolute mean deviation calculated can’t be subjected to further algebraic treatment. It can’t be used as an appropriate measure of dispersion.

15.5 Variance and standard deviation

15.5.1 Standard Deviation

This section involves the variance and standard deviation definition, formula and solved examples of the discrete and continuous frequency distribution.

A science test was taken by a class of students. The mean score of the test was 85% on calculation. The teacher found the standard deviation of other scores and noticed that a very small standard deviation exists, which suggests that most of the students scored very close to 85%.

15.5.2 Shortcut method to find Variance and standard deviation

This section deals with the simpler way of calculating the standard deviation with a few illustrations.

15.6 Analysis of Frequency Distributions

This section deals with the process of comparing the variability of two series having the same mean, coefficient of variation with few solved problems.

A Few Points on Chapter 15 Statistics

• The range is defined as the difference between the maximum and minimum values of the given data.
• If there exists a series with equal means, then the series with a lesser standard deviation is more consistent or less scattered.
• The addition or subtraction of a positive number to each data point of the data set will not affect the variance.

The solutions give substitute strategies and clarifications to solve problems which makes the student feel confident while taking the exam. Additionally, solving and practising new problems upgrades the understanding of Mathematical concepts among students. The solutions help with inquiries that a student has on the questions asked in the exam. The BYJU’S subject specialists who have written the NCERT Solutions for Class 11 Maths, according to the latest CBSE Syllabus 2023-24, have in-depth knowledge about the question paper setting and the marks distributed across the chapters.

Disclaimer – 

Dropped Topics – 

15.6 Analysis of Frequency Distribution
Ques. 6 (Miscellaneous Exercise) and last point in the Summary