# NCERT Solutions for Class 11 Maths Chapter 15 Statistics

NCERT Solutions for Class 11 Maths Chapter 15 Statistics: You have learnt the basic statistics like mean, mode and median known as "measures of the central tendency" in the previous class. In this chapter, you will learn important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range and many more. In NCERT solutions for class 11 maths chapter 15 statistics, you will get questions related to all the above topics. You will also learn some methods for finding a representative value for the given data. This value is known as the measure of central tendency. This chapter is important for the CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT, etc. As statistics deal with the collected data for specific purposes and in the digital world, everything that we do comes in the form of data. This is a very useful technique to interpret, analyze the data and make a conclusive decision based on the data. Solutions of NCERT for class 11 maths chapter 15 statistics will build your fundamentals of statistics which will be useful to learn advanced statistical techniques. There are 27 questions in 3 exercises in the NCERT textbook. All these questions are explained in CBSE NCERT solutions for class 11 maths chapter 15 statistics in a detailed manner. Check all NCERT solutions which will help you to understand the concepts in an easy way.
Topics of NCERT Grade 11 Maths Chapter-15 Statistics

15.1 Introduction

15.2 Measures of Dispersion

15.3 Range

15.4 Mean Deviation

15.5 Variance and Standard Deviation

15.6 Analysis of Frequency Distributions

The complete Solutions of NCERT Class 11 Mathematics Chapter 15 is provided below:

## NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.1

Mean () of the given data:

The respective absolute values of the deviations from mean,  are

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

Hence, the mean deviation about the mean is 3.

Mean () of the given data:

The respective absolute values of the deviations from mean,  are

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

Hence, the mean deviation about the mean is 8.4.

Question:3. Find the mean deviation about the median.

Number of observations, n = 12, which is even.

Arranging the values in ascending order:

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

Now, Median (M)

The respective absolute values of the deviations from median,  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

Hence, the mean deviation about the median is 2.33.

Question:4. Find the mean deviation about the median.

Number of observations, n = 10, which is even.

Arranging the values in ascending order:

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

Now, Median (M)

The respective absolute values of the deviations from median,  are

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

Hence, the mean deviation about the median is 7.

Question:5 Find the mean deviation about the mean.

 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 = 25 = 350 =158

Now, we calculate the absolute values of the deviations from mean,   and

= 158

Hence, the mean deviation about the mean is 6.32

Question:6.   Find the mean deviation about the mean.

 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 = 80 = 4000 =1280

Now, we calculate the absolute values of the deviations from mean,   and

= 1280

Hence, the mean deviation about the mean is 16

Question:7. Find the mean deviation about the median.

 5 8 8 2 16 7 6 14 0 0 9 2 16 2 4 10 2 18 3 6 12 2 20 5 10 15 6 26 8 48

Now, N = 26 which is even.

Median is the mean of  and  observations.

Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Therefore, Median, M

Now, we calculate the absolute values of the deviations from median,   and

= 84

Hence, the mean deviation about the median is 3.23

 15 3 3 13.5 40.5 21 5 8 7.5 37.5 27 6 14 1.5 9 30 7 21 1.5 10.5 35 8 29 6.5 52

Now, N = 30, which is even.

Median is the mean of  and  observations.

Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore, Median, M

Now, we calculate the absolute values of the deviations from median,   and

= 149.5

Hence, the mean deviation about the median is 5.1

Question:9. Find the mean deviation about the mean.

 Income per day in Rs Number of persons

 Income per day Number of  Persons Mid Points 0 -100 4 50 200 308 1232 100 -200 8 150 1200 208 1664 200-300 9 250 2250 108 972 300-400 10 350 3500 8 80 400-500 7 450 3150 92 644 500-600 5 550 2750 192 960 600-700 4 650 2600 292 1168 700-800 3 750 2250 392 1176 =50 =17900 =7896

Now, we calculate the absolute values of the deviations from mean,   and

= 7896

Hence, the mean deviation about the mean is 157.92

Question:10. Find the mean deviation about the mean.

 Height in cms Number of persond

 Height in cms Number of  Persons Mid Points 95 -105 9 100 900 25.3 227.7 105 -115 13 110 1430 15.3 198.9 115-125 26 120 3120 5.3 137.8 125-135 30 130 3900 4.7 141 135-145 12 140 1680 14.7 176.4 145-155 10 150 1500 24.7 247 =100 =12530 =1128.8

Now, we calculate the absolute values of the deviations from mean,   and

= 1128.8

Hence, the mean deviation about the mean is 11.29

 Marks Number of girls

 Marks Number of  Girls Cumulative Frequency c.f. Mid Points 0-10 6 6 5 22.85 137.1 10-20 8 14 15 12.85 102.8 20-30 14 28 25 2.85 39.9 30-40 16 44 35 7.15 114.4 40-50 4 48 45 17.15 68.6 50-60 2 50 55 27.15 54.3 =517.1

Now, N = 50, which is even.

The class interval containing  or  item is 20-30. Therefore, 20-30 is the median class.

We know,

Median

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median

Now, we calculate the absolute values of the deviations from median,   and

= 517.1

Hence, the mean deviation about the median is 10.34

 Age (in years) Number

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

 Age (in years) Number Cumulative Frequency c.f. Mid Points 15.5-20.5 5 5 18 20 100 20.5-25.5 6 11 23 15 90 25.5-30.5 12 23 28 10 120 30.5-35.5 14 37 33 5 70 35.5-40.5 26 63 38 0 0 40.5-45.5 12 75 43 5 60 45.5-50.5 16 91 48 10 160 50.5-55.5 9 100 53 15 135 =735

Now, N = 100, which is even.

The class interval containing  or  item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median

Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

Therefore, Median

Now, we calculate the absolute values of the deviations from median,   and

= 735

Hence, the mean deviation about the median is 7.35

NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.2

Mean () of the given data:

The respective values of the deviations from mean,  are

-3, -2, 1 3 4 -5 -1 3

Hence,  Mean = 9 and Variance = 9.25

First n natural numbers.

Mean () of first n natural numbers:

We know, Variance

We know that

Hence, Mean =  and Variance =

First 10 multiples of 3

First 10 multiples of 3 are:

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

Mean () of the above values:

The respective values of the deviations from mean,  are

-13.5, -10.5, -7.5, -4.5, -1.5, 1.5, 4.5, 7.5, 10.5, 13.5

Hence,  Mean = 16.5 and Variance = 74.25

 6 10 14 18 24 28 30 2 4 7 12 8 4 3

 6 2 12 -13 169 338 10 4 40 -9 81 324 14 7 98 -5 25 175 18 12 216 -1 1 12 24 8 192 5 25 200 28 4 112 9 81 324 30 3 90 13 169 363 = 40 = 760 =1736

We know, Variance,

Hence, Mean = 19 and Variance = 43.4

Question:​​​​​​​5. Find the mean and variance for each of the data.

 92 93 97 98 102 104 109 3 2 3 2 6 3 3

 92 3 276 -8 64 192 93 2 186 -7 49 98 97 3 291 -3 9 27 98 2 196 -2 4 8 102 6 612 2 4 24 104 3 312 4 16 48 109 3 327 9 81 243 = 22 = 2200 =640

We know, Variance,

Hence, Mean = 100 and Variance = 29.09

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

 60 61 62 63 64 65 66 67 68 2 1 12 29 25 12 10 4 5

Let the assumed mean, A = 64 and h = 1

 60 2 -4 16 -8 32 61 1 -3 9 -3 9 62 12 -2 4 -24 48 63 29 -1 1 -29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 =100 = 0 =286

Mean,

We know, Variance,

We know,  Standard Deviation =

Hence, Mean = 64 and Standard Deviation = 1.691

Question:​​​​​​​7.Find the mean and variance for the following frequency distributions.

 Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2

 Classes Frequency Mid point 0-30 2 15 30 -92 8464 16928 30-60 3 45 135 -62 3844 11532 60-90 5 75 375 -32 1024 5120 90-120 10 105 1050 2 4 40 120-150 3 135 405 28 784 2352 150-180 5 165 825 58 3364 16820 180-210 2 195 390 88 7744 15488 = N = 30 = 3210 =68280

We know, Variance,

Hence, Mean = 107 and Variance = 2276

Question:​​​​​​​8.  Find the mean and variance for the following frequency distributions.

 Classes 0-10 10-20 20-30 30-40 40-50 Frequencies 5 8 15 16 6

 Classes Frequency Mid-point 0-10 5 5 25 -22 484 2420 10-20 8 15 120 -12 144 1152 20-30 15 25 375 -2 4 60 30-40 16 35 560 8 64 1024 40-50 6 45 270 18 324 1944 = N = 50 = 1350 =6600

We know, Variance,

Hence, Mean = 27 and Variance = 132

 Height in cms 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115 No. of students 3 4 7 7 15 9 6 6 3

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

 Height in cms Frequency Midpoint 70-75 3 72.5 -4 16 -12 48 75-80 4 77.5 -3 9 -12 36 80-85 7 82.5 -2 4 -14 28 85-90 7 87.5 -1 1 -7 7 90-95 15 92.5 0 0 0 0 95-100 9 97.5 1 1 9 9 100-105 6 102.5 2 4 12 24 105-110 6 107.5 3 9 18 54 110-115 3 112.5 4 16 12 48 =N = 60 = 6 =254

Mean,

We know, Variance,

We know,  Standard Deviation =

Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

 Diameters 33-36 37-40 41-44 45-48 49-52 No. of circles 15 17 21 22 25

Calculate the standard deviation and mean diameter of the circles.
[ Hint First make the data continuous by making the classes as   and then proceed.]

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

 Diameters No. of circles Midpoint 32.5-36.5 15 34.5 -2 4 -30 60 36.5-40.5 17 38.5 -1 1 -17 17 40.5-44.5 21 42.5 0 0 0 0 44.5-48.5 22 46.5 1 1 22 22 48.5-52.5 25 50.5 2 4 50 100 =N = 100 = 25 =199

Mean,

We know, Variance,

We know,  Standard Deviation =

Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553

CBSE NCERT solutions for class 11 maths chapter 15 statistics-Exercise: 15.3

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

 Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group A 9 17 32 33 40 10 9 Group B 10 20 30 25 43 15 7

The group having a higher coefficient of variation will be more variable.

Let the assumed mean, A = 45 and h = 10

For Group A

 Marks Group A Midpoint 10-20 9 15 -3 9 -27 81 20-30 17 25 -2 4 -34 68 30-40 32 35 -1 1 -32 32 40-50 33 45 0 0 0 0 50-60 40 55 1 1 40 40 60-70 10 65 2 4 20 40 70-80 9 75 3 9 27 81 =N = 150 = -6 =342

Mean,

We know, Variance,

We know,  Standard Deviation =

Coefficient of variation =

C.V.(A) =

Similarly,

For Group B

 Marks Group A Midpoint 10-20 10 15 -3 9 -30 90 20-30 20 25 -2 4 -40 80 30-40 30 35 -1 1 -30 30 40-50 25 45 0 0 0 0 50-60 43 55 1 1 43 43 60-70 15 65 2 4 30 60 70-80 7 75 3 9 21 72 =N = 150 = -6 =375

Mean,

We know, Variance,

We know,  Standard Deviation =

Coefficient of variation =

C.V.(B) =

Since C.V.(B) > C.V.(A)

Therefore, Group B is more variable.

 X 35 54 52 53 56 58 52 50 51 49 Y 108 107 105 105 106 107 104 103 104 101

 X() Y() 35 108 1225 11664 54 107 2916 11449 52 105 2704 11025 53 105 2809 11025 56 106 8136 11236 58 107 3364 11449 52 104 2704 10816 50 103 2500 10609 51 104 2601 10816 49 101 2401 10201 =510 = 1050 =26360 =110290

For X,

Mean ,

Variance,

We know,  Standard Deviation =

C.V.(X) =

Similarly, For Y,

Mean ,

Variance,

We know,  Standard Deviation =

C.V.(Y) =

Since C.V.(Y) < C.V.(X)

Hence Y is more stable.

 Firm A Firm B No. of wage earners Mean of monthly wages Variance of the distribution of wages

Which firm A or B pays larger amount as monthly wages?

Given, Mean of monthly wages of firm A = 5253

Number of wage earners = 586

Total amount paid = 586 x 5253 = 30,78,258

Again, Mean of monthly wages of firm B = 5253

Number of wage earners = 648

Total amount paid = 648 x 5253 = 34,03,944

Hence firm B pays larger amount as monthly wages.

 Firm A Firm B No. of wage earners Mean of monthly wages Variance of the distribution of wages

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

Given, Variance of firm A = 100

Standard Deviation =

Again, Variance of firm B = 121

Standard Deviation =

Since , firm B has greater variability in individual wages.

 No. of goals scored 0 1 2 3 4 No. of matches 1 9 7 5 3

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