# NCERT Solutions For Class 11 Maths Chapter 15

## Ncert Solutions For Class 11 Maths Chapter 15 PDF Free Download

NCERT Solutions For Class 11 Maths Chapter 15 Statistics are given here in a detailed and an easy way to understand. These solutions for analysis of statistics class 11 will help the students to clear all their doubts instantly and let them know the best methods of solving these questions. These NCERT class 11 maths solutions for statistics (chapter 14) are available in pdf for free and any student can download in their computers and study offline as well.

The class 11 NCERT maths solutions for chapter 15 given here is created according to the latest guidelines of the Central Board of Secondary Education (CBSE). All the NCERT Solutions For Class 11 Maths given here are detailed and in easy to understand the simple steps.

## Class 11 Maths NCERT Solutions – Statistics Mean Median

At BYJU’S, students are also provided with notes, sample papers and previous year question papers to help them learn more effectively. NCERT solutions for Class 11 covers the important topics included in Maths chapter 15 Statistics, such as;

• Measures of Dispersion
• Range and Mean Deviation
• Mean Deviation for grouped and ungrouped data
• Variance and Standard Deviations
• Analysis of Frequency Distributions

Also, BYJU’S provides exemplar problems to practice more of different types of questions and become an excel in the topics. Students can practice exercise-wise solutions for Chapter 15, and can download the statistics class 11 notes as well in PDF format.

### NCERT Solutions Class 11 Maths Chapter 15 Exercises

Even for the statistics chapter, the complete exercise-wise solutions are given which can be understood by all the students and can help them to have a better understanding of the concepts. These solutions are formulated as per class 11 maths CBSE syllabus (2018-2019), prescribed by the board.

#### Exercise 15.1

Q1. Calculate the mean deviation about the mean for the given data

5, 8, 9, 10, 11, 13, 14, 18

Sol:

The given data is,

5, 8, 9, 10, 11, 13, 14, 18

Mean,

$\bar{x} = \frac{5 + 8 + 9 + 10 + 11 + 13 + 14 + 18}{8}$ $\bar{x} = \frac{88}{8}$ $\bar{x} = 11$

The deviations of the respective observations from the mean $\bar{x}$,

$x_{i} – \bar{x}$ are -6, -3, -2, -1, 0, 2, 3, 7

The absolute values of the deviations,

$\left |x_{i} – \bar{x} \right |$ are 6, 3, 2, 1, 0 , 2, 3, 7

The required mean deviation about the mean is,

$M.D. \;\left (\bar{x} \right ) = \frac{\sum_{i = 1}^{8} \left | x_{i} – \bar{x} \right |}{8}$ $M.D. \;\left (\bar{x} \right ) = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8}$ $M.D. \;\left (\bar{x} \right ) = \frac{24}{8}$ = 3

Q2. Calculate the mean deviation about the mean for the given data

39, 71, 49, 41, 43, 56, 64, 47, 55, 45

Sol:

The given data is,

39, 71, 49, 41, 43, 56, 64, 47, 55, 45

Mean,

$\bar{x} = \frac{39 + 71 + 49 + 41 + 43 + 56 + 64 + 47 + 55 + 45}{10}$ $\bar{x} = \frac{510}{10}$ $\bar{x} = 51$

The deviations of the respective observations from the mean $\bar{x}$,

$x_{i} – \bar{x}$ are -12, 20, -2, -10, -8, 5, 13, -4, 4, -6.

The absolute values of the deviations,

$\left |x_{i} – \bar{x} \right |$ are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6.

The required mean deviation about the mean is,

$M.D. \;\left (\bar{x} \right ) = \frac{\sum_{i = 1}^{10} \left | x_{i} – \bar{x} \right |}{10}$ $M.D. \;\left (\bar{x} \right ) = \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10}$ $M.D. \;\left (\bar{x} \right ) = \frac{84}{10}$ = 8.4

Q3. Calculate the mean deviation about the median for the given data:

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

Sol:

The given data is,

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

There are 12 observations which is even number.

Arrange the data in ascending order:

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

Median,

$M = \frac{\left (\frac{12}{2} \right )^{th} observation + \left ( \frac{12}{2} + 1 \right )^{th} observation}{2}$

= $\frac{6 ^{th}\; observation + 7^{th}\; observation}{2}$

= $\frac{13 + 14}{2}$

= $\frac{27}{2}$

= 13.5

The deviations of the respective observations from the median $\bar{x}$,

$x_{i} – M$ 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

The absolute values of the deviations,

$\left |x_{i} – M \right |$ 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

The required mean deviation about the median is,

$M.D. \;\left (M\right ) = \frac{\sum_{i = 1}^{12} \left | x_{i} – M \right |}{12}$

= $\frac{3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5}{12}$

= $\frac{28}{12}$

= 2.33

Q4. Calculate the mean deviation about the median for the given data:

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

Sol:

The given data is,

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

There are 10 observations which is even number.

Arrange the data in ascending order:

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

Median,

$M = \frac{\left (\frac{10}{2} \right )^{th} observation + \left ( \frac{10}{2} + 1 \right )^{th} observation}{2}$

= $\frac{5 ^{th}\; observation + 6^{th}\; observation}{2}$

= $\frac{46 + 49}{2}$

= $\frac{95}{2}$

= 47.5

The deviations of the respective observations from the median $\bar{x}$,

$x_{i} – M$ are –11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations,

$\left |x_{i} – M \right |$ are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The required mean deviation about the median is,

$M.D. \;\left (M\right ) = \frac{\sum_{i = 1}^{10} \left | x_{i} – M \right |}{10}$

= $\frac{11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5}{10}$

= $\frac{70}{10}$

= 7

Q5. Calculate the mean deviation about the mean for the given data

 $x_{i}$ 5 10 15 20 25 $f_{i}$ 7 4 6 3 5

Sol:

 $x_{i}$ $x_{i}$ $x_{i}$ $|x_{i} – \bar{x}|$ $f_{i} |x_{i} – \bar{x}|$ 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

$N = \sum_{i = 1}^{5}f_{i} = 25$

$\sum_{i = 1}^{5}f_{i}x_{i} = 350$

Therefore,

$\bar{x} = \frac{1}{N}\sum_{i = 1}^{5}f_{i}x_{i}$

= $\frac{1}{25} \times 350$

= 14

$MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – \bar{x} \right |$

= $\frac{1}{25} \times 158$

= 6.32

Q6. Calculate the mean deviation about the mean for the given data

 $x_{i}$ 10 10 15 20 25 $f_{i}$ 7 4 6 3 5

Sol:

 $x_{i}$ $x_{i}$ $x_{i}$ $|x_{i} – \bar{x}|$ $f_{i} |x_{i} – \bar{x}|$ 10 4 35 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

$N = \sum_{i = 1}^{5}f_{i} = 80$

$\sum_{i = 1}^{5}f_{i}x_{i} = 4000$

Therefore,

$\bar{x} = \frac{1}{N}\sum_{i = 1}^{5}f_{i}x_{i}$

= $\frac{1}{80} \times 4000$

= 50

$MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – \bar{x} \right |$

= $\frac{1}{80} \times 1280$

= 16

Q7. Calculate the mean deviation about the median for the given data

 $x_{i}$ 10 10 15 20 25 $f_{i}$ 7 4 6 3 5
 $x_{i}$ 5 7 9 10 12 15 $f_{i}$ 8 6 2 2 2 6

Sol:

The observations given are in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:

 $x_{i}$ $f_{i}$ c.f. 5 8 8 7 6 14 9 2 16 10 2 18 12 2 20 15 6 26

Here, N = 26, which is even.

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

$Median = \frac{13^{th}\; observation + 14^{th} observation}{2}$

= $\frac{7 + 7}{2}$

= 7

The absolute values of the deviations from median,

 $|x_{i} – M|$ 2 0 2 3 5 8 $f_{i}$ 8 6 2 2 2 6 $f_{i}|x_{i} – M|$ 16 0 4 6 10 48

$\sum_{i = 1}^{6}f_{i} = 26$

$\sum_{i = 1}^{6}f_{i}|x_{i} – M| = 84$

Therefore,

$MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – M\right |$

= $\frac{1}{26} \times 84$

= 3.23

Q8. Calculate the mean deviation about the median for the given data

 $x_{i}$ 15 21 27 30 35 $f_{i}$ 3 5 6 7 8

Sol:

The observations given are in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:

 $x_{i}$ $f_{i}$ c.f. 15 3 3 21 5 8 27 6 14 30 7 21 35 8 29

Here, N = 29, which is odd.

$Median = \left (\frac{29 + 1}{2} \right )^{th} observation$

= 15th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore,

Median = 30

The absolute values of the deviations from median,

 $|x_{i} – M|$ 15 9 3 0 5 $f_{i}$ 3 5 6 7 8 $f_{i}|x_{i} – M|$ 45 45 18 0 40

$\sum_{i = 1}^{5}f_{i} = 29$

$\sum_{i = 1}^{5}f_{i}|x_{i} – M| = 148$

Therefore,

$MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – M\right |$

= $\frac{1}{29} \times 148$

= 5.1

Q9. Calculate the mean deviation about the mean for the given data:

 Income per day Number of persons 0 – 100 4 100 – 200 8 200 – 300 9 300 – 400 10 400 – 500 7 500 – 600 5 600 – 700 4 700 – 800 3

Sol:

The given table is formed.

 Income per day Number of persons $f_{i}$ Mid – point $x_{i}$ $f_{i}$ $x_{i}$ $|x_{i} – \bar{x}|$ $f_{i}|x_{i} – \bar{x}|$ 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

$N = \sum_{i = 1}^{8}f_{i} = 50$

$\sum_{i = 1}^{8}f_{i}x_{i} = 17900$

Therefore,

$\bar{x} = \frac{1}{N}\sum_{i = 1}^{8}f_{i}x_{i}$

= $\frac{1}{50} \times 17900$

= 358

$MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{8}f_{i}\left |x_{i} – \bar{x} \right |$

= $\frac{1}{50} \times 7896$

= 175.92

Q10. Calculate the mean deviation about the mean for the given data:

 Height in cm Number of boys 95 – 105 9 105 – 115 13 115 – 125 26 125 – 135 30 135 – 145 12 145 – 155 10

Sol:

The given table is obtained:

 Height in cm Number of boys $f_{i}$ Mid – point $x_{i}$ $f_{i}$ $x_{i}$ $|x_{i} – \bar{x}|$ $f_{i}|x_{i} – \bar{x}|$ 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

$N = \sum_{i = 1}^{6}f_{i} = 100$

$\sum_{i = 1}^{6}f_{i}x_{i} = 12530$

Therefore,

$\bar{x} = \frac{1}{N}\sum_{i = 1}^{6}f_{i}x_{i}$

= $\frac{1}{100} \times 12530$

= 125.3

$MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – \bar{x} \right |$

= $\frac{1}{100} \times 1128.8$

= 11.28

Q11. Calculate the mean deviation about the median for the given data:

 Marks Number of girls 0 – 10 6 10 – 20 8 20 – 30 14 30 – 40 16 40 – 50 4 50 – 60 2

Sol:

The given table is obtained:

 Marks Number of girls $f_{i}$ Cumulative frequency (c.f.) Mid – point $x_{i}$ $|x_{i} – M|$ $f_{i}|x_{i} – \bar{x}|$ 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 50 517.1

The class interval containing $\left ( \frac{N}{2} \right )^{th}$ or 25th item is 20 – 30.

Therefore, 20 -30 is the median class.

Now,

$Median = l + \frac{\frac{N}{2} – C}{f} \times h$

Where, l = 20

C = 14

f = 14

h = 10

N = 50

$Median = 20 + \frac{25 – 14}{14} \times 10$

= $20 + \frac{110}{14}$

= 20 + 7.85

= 27.85

Mean deviation about the median is,

$MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – M\right |$

= $\frac{1}{50} \times 517.1$

= 10.34

Q12.  Calculate the mean deviation about the median for the given data:

 Age Number 16 – 20 5 21 – 25 6 26 – 30 12 31 – 35 14 36 – 40 26 41 – 45 12 46 – 50 16 51 – 55 9

Sol:

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is as given below:

 Age Number $f_{i}$ Cumulative frequency (c.f.) Mid – point $x_{i}$ $|x_{i} – M|$ $f_{i}|x_{i} – \bar{x}|$ 16 – 20 5 5 18 20 100 21 – 25 6 11 23 15 90 26 – 30 12 23 28 10 120 31 – 35 14 37 33 5 70 36 – 40 26 63 38 0 0 41 – 45 12 75 43 5 60 46 – 50 16 91 48 10 160 51 – 55 9 100 53 15 135 100

The class interval containing $\left ( \frac{N}{2} \right )^{th}$ or 50th item is 35.5 – 40.5. Therefore, 35.5 – 40.5 is the median class.

Now,

$Median = l + \frac{\frac{N}{2} – C}{f} \times h$

Where, l = 35.5

C = 37

f = 26

h = 5

N = 100

$Median = 35.5 + \frac{50 – 37}{26} \times 5$

= $35.5 + \frac{13 × 5}{26}$

= 35.5 + 2.5

= 38

Mean deviation about the median is,

$MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{8}f_{i}\left |x_{i} – M\right |$

= $\frac{1}{100} \times 735$

= 7.35

#### Exercise 15.2

Q1. Calculate mean and variance for the given data 6, 7, 10, 12, 13, 4, 8, 12

Sol:

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

Mean,

$\bar{x} = \frac{\sum_{i = 1}^{8}x_{i}}{n}$

= $\frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8}$

= $\frac{72}{8}$

= 9

The given table is obtained,

 $x_{i}$ $\left ( x_{i} – \bar{x} \right )$ $\left ( x_{i} – \bar{x} \right )^{2}$ 6 -3 9 7 -2 4 10 -1 1 12 3 9 13 4 16 4 -5 25 8 -1 1 12 3 9 74

Variance,

$\left ( \sigma ^{2} \right ) = \frac{1}{n}\sum_{i = 1}^{8}\left ( x_{i} – \bar{x} \right )^{2}$

= $\frac{1}{8} \times 74$

= 9.25

Q2. Calculate the mean and variance for the first n natural numbers.

Sol:

The mean of first n natural number is,

$Mean = \frac{Sum \; of \; all \; observations}{Number \; of \; observations}$ $Mean = \frac{\frac{n\left ( n + 1 \right )}{2}}{n}$

= $\frac{ n + 1 }{2}$

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{1}{n}\sum_{i = 1}^{n}\left ( x_{i} – \bar{x} \right )^{2}$

= $\frac{1}{n}\sum_{i = 1}^{n}\left [ x_{i} – \left ( \frac{n + 1}{2} \right ) \right ]^{2}$

= $\frac{1}{n}\sum_{i = 1}^{n} x_{i}^{2} – \frac{1}{n}\sum_{i = 1}^{n}2\left ( \frac{n + 1}{2} \right )x_{i} + \frac{1}{n}\sum_{i = 1}^{n}\left ( \frac{n + 1}{2} \right )^{2}$

= $\frac{1}{n}\frac{n\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \left ( \frac{n + 1}{n} \right )\left [ \frac{n\left ( n + 1 \right )}{2} \right ] + \frac{\left ( n + 1 \right )^{2}}{4n} \times n$

= $\frac{\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \frac{\left (n + 1 \right )^{2}}{2} + \frac{\left ( n + 1 \right )^{2}}{4}$

= $\frac{\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \frac{\left ( n + 1 \right )^{2}}{4}$

= $\left ( n + 1 \right )\left [\frac{ 4n + 2 – 3n – 3 }{12}\right ]$

= $\frac{\left ( n + 1 \right )\left ( n – 1 \right )}{12}$

= $\frac{n^{2} – 1}{12}$

Q3. Calculate the mean and variance for the first 10 multiples of 3.

Sol:

The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Number of observations,

n = 10

Mean,

$\bar{x} = \frac{\sum_{i = 1}^{10}}{10}$

= $\frac{165}{10}$

= 16.5

The given table is obtained,

 $x_{i}$ $(x_{i} – \bar{x})$ $(x_{i} – \bar{x})^{2}$ 3 -13.5 182.25 6 -10.5 110.25 9 -7.5 56.25 12 -4.5 20.25 15 -1.5 2.25 18 1.5 2.25 21 4.5 20.25 24 7.5 56.25 27 10.5 110.25 30 13.5 182.25 742.5

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{1}{n}\sum_{i = 1}^{10}\left ( x_{i} – \bar{x} \right )^{2}$

= $\frac{1}{10} \times 742.5$

= 74.25

Q4. Calculate the mean and variance for the data

 $x_{i}$ 6 10 14 18 24 28 30 $f_{i}$ 2 4 7 12 8 4 3

Sol:

The data obtained is given in the tabular form:

 $x_{i}$ $x_{i}$ $f_{i}x_{i}$ $x_{i} – \bar{x}$ $(x_{i}-\bar{x})^{2}$ $f_{i}(x_{i}-\bar{x})^{2}$ 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 11 121 363 40 760 1736

Here,

N = 40

$\sum_{i = 1}^{7}f_{i}x_{i}$ = 760

$\bar{x} = \frac{\sum_{i = 1}^{7}f_{i}x_{i}}{N}$

= $\frac{760}{40}$

= 19

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{1}{n}\sum_{i = 1}^{7}\left ( x_{i} – \bar{x} \right )^{2}$

= $\frac{1}{40} \times 1736$

= 43.4

Q5. Calculate the mean and variance for the data

 $x_{i}$ 92 93 97 98 102 104 109 $f_{i}$ 3 2 3 2 6 3 3

Sol:

The data obtained is given in the tabular form:

 $x_{i}$ $f_{i}$ $f_{i}x_{i}$ $x_{i} – \bar{x}$ $(x_{i}-\bar{x})^{2}$ $f_{i}(x_{i}-\bar{x})^{2}$ 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

Here,

N = 22

$\sum_{i = 1}^{7}f_{i}x_{i}$ = 2200

$\bar{x} = \frac{\sum_{i = 1}^{7}f_{i}x_{i}}{N}$

= $\frac{2200}{22}$

= 100

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{1}{n}\sum_{i = 1}^{7}\left ( x_{i} – \bar{x} \right )^{2}$

= $\frac{1}{22} \times 640$

= 29.09

Q6. Calculate the mean and standard deviation using short-cut method.

 $x_{i}$ 60 61 62 63 64 65 66 67 68 $f_{i}$ 2 1 12 29 25 12 10 4 5

Sol:
The data is obtained in tabular form as follows.

 $x_{i}$ $f_{i}$ $f_{i}=\frac{x_{i}-64}{1}$ $y_{i}^{2}$ $f_{i}y_{i}$ $f_{i}y_{i}^{2}$ 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 80 100 220 0 286

Mean,

$\bar{x} = A\frac{\sum_{i = 1}^{9}f_{i}y_{i}}{N} \times h$

= $64 + \frac{0}{100} \times 1$

= 64 + 0

= 64

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{9}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{9}f_{i}y_{i} \right )^{2}\right ]$

= $\frac{1}{100^{2}}[ 100 \times 286 – 0]$

= 2.86

Standard deviation,

$\sigma = \sqrt{2.86}$

= 1.69

Q7. Calculate the mean and variance for the given frequency distribution.

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

Sol:

 Class $f_{i}$ ${x_{i}}$ $y_{i}=\frac{x_{i}-105}{30}$ $y_{i}^{2}$ $f_{i}y_{i}$ $f_{i}y_{i}^{2}$ 0 – 30 2 15 -3 9 -6 18 30 – 60 3 45 -2 4 -6 12 60 – 90 5 75 -1 1 -5 5 90 – 120 10 105 0 0 0 0 120 – 150 3 135 1 1 3 3 150 – 180 5 165 2 4 10 20 180 – 210 2 195 3 9 6 18 30 2 76

Mean,

$\bar{x}$

= $A + \frac{\sum_{i = 1}^{7}f_{i}y_{i}}{N} \times h$

= $105 + \frac{2}{30} \times 30$

= 105 + 2

= 107

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{7}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{7}f_{i}y_{i} \right )^{2}\right ]$

= $\frac{30^{2}}{30^{2}}\left [ 30 \times 76 – \left ( 2 \right )^{2}\right ]$

= 2280 – 4

= 2276

Q8. Calculate the mean and variance for the following frequency distribution.

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

Sol:

 Class Frequency $f_{i}$ Mid – point $x_{i}$ $y_{i}=\frac{x_{i}-25}{10}$ $y_{i}^{2}$ $f_{i}y_{i}$ $f_{i}y_{i}^{2}$ 0 – 10 5 5 -2 4 -10 20 10 – 20 8 15 -1 1 -8 8 20 – 30 15 25 0 0 0 0 30 – 40 16 35 1 1 16 16 40 – 50 6 45 2 4 12 24 50 10 68

Mean,

$\bar{x}$

= $A + \frac{\sum_{i = 1}^{5}f_{i}y_{i}}{N} \times h$

= $25 + \frac{10}{50} \times 10$

= 25 + 2

= 27

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{5}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{5}f_{i}y_{i} \right )^{2}\right ]$

= $\frac{10^{2}}{50^{2}}\left [ 50 \times 68 – \left ( 10 \right )^{2}\right ]$

= $\frac{1}{25}\left [ 3400 – 100 \right ]$

= $\frac{3300}{25}$

= 132

Q9. Calculate the mean, variance and standard deviation using short – cut method:

 Height in cm Number of children 70 – 75 3 75 – 80 4 80 – 85 7 85 – 90 7 90 – 95 15 95 – 100 9 100 – 105 6 105 – 110 6 110 – 115 3

Sol:

 Class $f_{i}$ $x_{i}$ $y_{i}=\frac{x_{i}-92.5}{5}$ $y_{i}^{2}$ $f_{i}y_{i}$ $f_{i}y_{i}^{2}$ 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 60 6 254

Mean,

$\bar{x}$

= $A + \frac{\sum_{i = 1}^{9}f_{i}y_{i}}{N} \times h$

= $92.5 + \frac{6}{60} \times 5$

= 92.5 + 0.5

= 93

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{9}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{9}f_{i}y_{i} \right )^{2}\right ]$

= $\frac{5^{2}}{60^{2}}\left [ 60 \times 254 – \left ( 6 \right )^{2}\right ]$

= $\frac{25}{3600}\left [ 155204 \right ]$

= 105.58

Standard deviation,

$\left (\sigma \right )$

= $\sqrt{105.58}$

= 10.27

Q10. The diameters of circles (in mm) drawn in a design are given below:

 Diameters Number of Children 33 – 36 15 37 – 40 17 41 – 44 21 45 – 48 22 49 – 52 25

Sol:

 Class $f_{i}$ $x_{i}$ $y_{i}=\frac{x_{i}-42.5}{4}$ $y_{i}^{2}$ $f_{i}y_{i}$ $f_{i}y_{i}^{2}$ 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 50 100 100 25 199

N = 100

h = 4

Let us assume that mean A be 42.5

Mean,

$\bar{x}$

= $A + \frac{\sum_{i = 1}^{5}f_{i}y_{i}}{N} \times h$

= $42.5 + \frac{25}{100} \times 4$

= 43.5

Variance,

$\left ( \sigma ^{2} \right )$

= $\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{5}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{5}f_{i}y_{i} \right )^{2}\right ]$

= $\frac{4^{2}}{100^{2}}\left [ 100 \times 199 – \left ( 25 \right )^{2}\right ]$

= $\frac{16}{10000}\left [ 19900 – 625 \right]$

= $\frac{16}{10000} \times 19275$

= 30.84

Standard deviation,

$\left (\sigma \right )$

= $\sqrt{30.84}$

= 5.55

#### Exercise 15.3

Q1. 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

Sol:

The standard deviation of group A is calculated as given in the table below.

 Marks $f_{i}$ $x_{i}$ $y_{i}=\frac{x_{i}-45}{10}$ $y_{i}^{2}$ $f_{i}y_{i}$ $f_{i}y_{i}^{2}$ 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 81 150 -6 342

N = 150

h = 10

A = 45

Mean,

$\bar{x}$

= $A + \frac{\sum_{i = 1}^{7}x_{i}}{N} \times h$

= $45 + \frac{(-6)}{150} \times 10$

= 45 – 0.4

= 44.6

Variance,

$\left ( \sigma _{1}^{2} \right )$

= $\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{7}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{7}f_{i}y_{i} \right )^{2}\right]$

= $\frac{10^{2}}{110^{2}}\left [ 150 \times 342 – \left ( -6 \right )^{2}\right ]$

= $\frac{100}{22500}\left [ 51264 \right]$

= $\frac{1}{225} \times 51264$

= 227.84

Standard deviation,

$\left (\sigma _{1}\right )$

= $\sqrt{227.84}$

= 15.09

The standard deviation of group A is calculated as given in the table below.

 Marks $f_{i}$ $x_{i}$ $y_{i}=\frac{x_{i}-45}{10}$ $y_{i}^{2}$ $f_{i}y_{i}$ $f_{i}y_{i}^{2}$ 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 63 150 -6 366

N = 150

h = 10

A = 45

Mean,

$\bar{x}$

= $A + \frac{\sum_{i = 1}^{7}x_{i}}{N} \times h$

= $45 + \frac{(-6)}{150} \times 10$

= 45 – 0.4

= 44.6

Variance,

$\left ( \sigma _{2}^{2} \right )$

= $\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{7}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{7}f_{i}y_{i} \right )^{2}\right]$

= $\frac{10^{2}}{110^{2}}\left [ 150 \times 366 – \left ( -6 \right )^{2}\right ]$

= $\frac{100}{22500}\left [ 54864 \right]$

= $\frac{1}{225} \times 54864$

= 243.84

Standard deviation,

$\left (\sigma _{1}\right )$

= $\sqrt{243.84}$

= 15.61

As the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group B has more variability in the marks.

Q2. From the prices of shares X and Y below, find out which is more stable in value.

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

Sol:

The prices of the shares X are 35, 54, 52, 53, 56, 58, 52, 50, 51, and 49.

Here, the number of observations, N = 10

Mean,

$\bar{x}$

= $\frac{1}{N}\sum_{i = 1}^{10}x_{i}$

= $\frac{1}{10} \times 510$

= 51

The following table is obtained corresponding to shares X.

 $x_{i}$ $(x_{i}-\bar{x})$ $(x_{i}-\bar{x})^{2}$ 35 -16 256 54 3 9 52 1 1 53 2 4 56 5 25 58 7 49 52 1 1 50 -1 1 51 0 0 49 -2 4 350

Variance,

$\left ( \sigma _{1}^{2} \right )$

= $\frac{1}{N}\sum_{i = 1}^{10}\left ( x_{i} – \bar{x} \right )^{2}$

= $\frac{1}{10} \times 350$

= 35

Standard deviation,

$\left (\sigma _{1}\right )$

= $\sqrt{35}$

= 5.91

C.V.

$\frac{\sigma _{1}}{x} \times 100$

= $\frac{5.91}{51} \times 100$

= 11.58

The prices of share Y are 108, 107, 105, 105, 106, 107, 104, 103, 104, and 101.

Mean,

$\bar{y}$

= $\frac{1}{N}\sum_{i = 1}^{10}y_{i}$

= $\frac{1}{10} \times 1050$

= 105

The following table is obtained corresponding to shares Y.

 $y_{i}$ $(y_{i}-\bar{y})$ $(y_{i}-\bar{y})^{2}$ 108 3 9 107 2 4 105 0 0 105 0 0 106 1 1 107 2 4 104 -1 1 103 -2 4 104 -1 1 101 -4 16 40

Variance,

$\left ( \sigma _{2}^{2} \right )$

= $\frac{1}{N}\sum_{i = 1}^{10}\left ( y_{i} – \bar{y} \right )^{2}$

= $\frac{1}{10} \times 40$

= 4

Standard deviation,

$\left (\sigma _{2}\right )$

= $\sqrt{4}$

= 2

C.V.

$\frac{\sigma _{2}}{y} \times 100$

= $\frac{2}{105} \times 100$

= 1.9

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Therefore, the prices of shares Y are more stable than the prices of shares X.

Q3. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

 Firm A Firm B No. of wage earners 586 648 Mean of monthly wages Rs. 5253 Rs. 5253 Variance of the distribution of wages 100 121

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

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

Sol:

(i)

Monthly wages of firm A = Rs. 5253

Number of wage earners in firm A = 586

Therefore,

Total amount paid = Rs. 5253 × 586

Monthly wages of firm B = Rs. 5253

Number of wage earners in firm B = 648

Therefore,

Total amount paid = Rs. 5253 × 648

Hence, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii)

Variance of the distribution of wages in firm A,

$A\left ( \sigma _{1}^{2} \right )$ = 100

Therefore,

Standard deviation of the distribution of wages in firm A,

$A\left ( \sigma _{1} \right ) = \sqrt{100}$ = 10

Variance of the distribution of wages in firm B,

$B\left ( \sigma _{2}^{2} \right )$ = 121

Therefore,

Standard deviation of the distribution of wages in firm B,

$B\left ( \sigma _{2} \right ) = \sqrt{121}$ = 11

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.

Therefore, firm B has greater variability in the individual wages.

Q4. The following is the record of goals scored by team A in a football session:

 Number of goals scored 0 1 2 3 4 Number of matches 1 9 7 5 3

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

Sol:
The mean and the standard deviation of goals scored by team A are calculated as follows.

 No. of goals scored No. of matches $f_{i}x_{i}$ $x_{i}^{2}$ $f_{i}x_{i}^{2}$ 0 1 0 0 0 1 9 9 1 9 2 7 14 4 28 3 5 15 9 45 4 3 12 16 48 23 50 130

Mean,

$\frac{\sum_{i = 1}^{5}f_{i}x_{i}}{\sum_{i = 1}^{5}f_{i}}$

= $\frac{50}{25}$

= 2

Therefore, the mean of both the teams is same.

Standard deviation,

$\sigma$

= $\frac{1}{N}\sqrt{N\sum f_{i}x_{i}^{2} – \left ( \sum f_{i}x_{i} \right )^{2}}$

= $\frac{1}{25}\sqrt{25 \times 130 – \left ( 50 \right )^{2}}$

= $\frac{1}{25}\sqrt{750}$

= $\frac{1}{25} \times 27.38$

= 1.09

The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same that is 2. Therefore, the team with lower standard deviation will be more consistent.

Therefore, team A is more consistent than team B.

Q5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are as given below:

$\sum_{i = 1}^{50}x_{i} = 212$

$\sum_{i = 1}^{50}x_{i}^{2} = 902.8$

$\sum_{i = 1}^{50}y_{i} = 261$

$\sum_{i = 1}^{50}y_{i}^{2} = 1457.6$

Which is more varying, the length or weight?

Sol:

Length

$\sum_{i = 1}^{50}x_{i} = 212$ $\sum_{i = 1}^{50}x_{i}^{2} = 902.8$

Here, N = 50

Mean,

$\bar{x}$

= $\frac{\sum_{i = 1}^{50}y_{i}}{N}$

= $\frac{212}{50}$

= 4.24

Variance,

$\sigma _{1}^{2}$

= $\frac{1}{N}\sum_{i = 1}^{50}\left ( x_{i} – \bar{x} \right )^{2}$

= $\frac{1}{50}\sum_{i = 1}^{50}\left ( x_{i} – 4.24 \right )^{2}$

= $\frac{1}{50}\sum_{i = 1}^{50} \left [x_{i}^{2} – 8.48x_{i} + 17.97 \right ]$

= $\frac{1}{50} \left [\sum_{i = 1}^{50}x_{i}^{2} – 8.48\sum_{i = 1}^{50}x_{i} + 17.97 \times 50 \right ]$

= $\frac{1}{50} \left [902.8 – 8.48 \times \left (212 \right ) + 898.5 \right ]$

= $\frac{1}{50} \left [1801.3 – 1797.76 \right ]$

= $\frac{1}{50} \times 3.54$

= 0.07

Standard deviation,

$\sigma _{i}\left ( Length \right )$

= $\sqrt{0.07}$

= 0.26

C.V. (Length)

= $\frac{Standard \; deviation}{Mean} \times 100$

= $\frac{0.26}{4.24} \times 100$

= 6.13

Weight

$\sum_{i = 1}^{50}y_{i} = 261$ $\sum_{i = 1}^{50}y_{i}^{2} = 1457.6$

Mean,

$\bar{y}$

= $\frac{\sum_{i = 1}^{50}y_{i}}{N}$

= $\frac{261}{50}$

= 5.22

Variance,

$\sigma _{2}^{2}$

= $\frac{1}{N}\sum_{i = 1}^{50}\left ( y_{i} – \bar{y} \right )^{2}$

= $\frac{1}{50}\sum_{i = 1}^{50}\left ( y_{i} – 5.22 \right )^{2}$

= $\frac{1}{50}\sum_{i = 1}^{50} \left [y_{i}^{2} – 10.44 y_{i} + 27.24 \right ]$

= $\frac{1}{50} \left [\sum_{i = 1}^{50}y_{i}^{2} – 10.44 \sum_{i = 1}^{50}y_{i} + 27.24 \times 50 \right ]$

= $\frac{1}{50} \left [1457.6 – 10.44 \times \left (261 \right ) + 1362 \right ]$

= $\frac{1}{50} \left [2819.6 – 2724.84 \right ]$

= $\frac{1}{50} \times 94.76$

= 1.89

Standard deviation,

$\sigma _{2}\left ( Weight \right )$

= $\sqrt{1.89}$

= 1.37

C.V. (Weight)

= $\frac{Standard \; deviation}{Mean} \times 100$

= $\frac{1.37}{5.22} \times 100$

= 26.24

Therefore, C.V. of weights is greater than the C.V. of lengths. Hence, weights vary more than the lengths.

In the statistics chapter of NCERT book for class 11 maths, several important concepts related to statistics is introduced. This chapter is extremely crucial as the concepts of statistics are used in higher classes we well and also at your graduate level.

In this chapter, Statistics and all the subtopics are explained in an elaborated manner. All these topics require the students to understand the concepts in-depth, thoroughly and remember the formulas properly.

Several examples are included in the statistics chapter to help the students understand the methodologies for solving the respective questions. Students are also required to solve the different exercise problems that are given in the textbook, to develop their problem-solving and logical abilities.

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