# 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

$\small 4,7,8,9,10,12,13,17$

Mean ($\overline{x}$) of the given data:

$\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10$

The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are

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

$\therefore$  $\sum_{i=1}^{8}|x_i - 10| = 24$

$\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|$

$= \frac{24}{8} = 3$

Hence, the mean deviation about the mean is 3.

$\small 38,70,48,40,42,55,63,46,54,44$

Mean ($\overline{x}$) of the given data:

$\\ \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{38+70+48+40+42+55+63+46+54+44}{10} \\ = \frac{500}{10} = 50$

The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are

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

$\therefore$  $\sum_{i=1}^{8}|x_i - 50| = 84$

$\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|$

$= \frac{84}{10} = 8.4$

Hence, the mean deviation about the mean is 8.4.

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

$\small 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$

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)

$\\ = \frac{(\frac{12}{2})^{th} observation + (\frac{12}{2}+ 1)^{th} observation}{2} \\ = \frac{13 + 14}{2} = \frac{27}{2}= 13.5$

The respective absolute values of the deviations from median, $|x_i - M|$ 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

$\therefore$  $\sum_{i=1}^{8}|x_i - 13.5| = 28$

$\therefore$ $M.D.(M) = \frac{1}{12}\sum_{i=1}^{n}|x_i - M|$

$= \frac{28}{12} = 2.33$

Hence, the mean deviation about the median is 2.33.

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

$\small 36, 72, 46, 42, 60, 45, 53, 46, 51, 49$

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)

$\\ = \frac{(\frac{10}{2})^{th} observation + (\frac{10}{2}+ 1)^{th} observation}{2} \\ = \frac{46 + 49}{2} = \frac{95}{2}= 47.5$

The respective absolute values of the deviations from median, $|x_i - M|$ are

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

$\therefore$  $\sum_{i=1}^{8}|x_i - 47.5| = 70$

$\therefore$ $M.D.(M) = \frac{1}{10}\sum_{i=1}^{n}|x_i - M|$

$= \frac{70}{10} = 7$

Hence, the mean deviation about the median is 7.

Question:5 Find the mean deviation about the mean.

$\small \\x_i\hspace{1cm}5\hspace{1cm}10\hspace{1cm}15\hspace{1cm}20\hspace{1cm}25\\\small f_i\hspace{1cm}7\hspace{1cm}4\hspace{1.15cm}6\hspace{1.22cm}3\hspace{1.3cm}5$

 $x_i$ $f_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 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 $\sum{f_i}$ = 25 $\sum f_ix_i$ = 350 $\sum f_i|x_i - \overline{x}|$ =158

$N = \sum_{i=1}^{5}{f_i} = 25 ; \sum_{i=1}^{5}{f_ix_i} = 350$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{350}{12} = 14$

Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$  and

$\sum f_i|x_i - \overline{x}|$ = 158

$\therefore$ $M.D.(\overline{x}) = \frac{1}{25}\sum_{i=1}^{n}f_i|x_i - \overline{x}|$

$= \frac{158}{25} = 6.32$

Hence, the mean deviation about the mean is 6.32

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

$\small \\x_i\hspace{1cm}10\hspace{1cm}30\hspace{1cm}50\hspace{1cm}70\hspace{1cm}90\\\small f_i\hspace{1cm}4\hspace{1.1cm}24\hspace{1.1cm}28\hspace{1.1cm}16\hspace{1.2cm}8$

 $x_i$ $f_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 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 $\sum{f_i}$ = 80 $\sum f_ix_i$ = 4000 $\sum f_i|x_i - \overline{x}|$ =1280

$N = \sum_{i=1}^{5}{f_i} = 80 ; \sum_{i=1}^{5}{f_ix_i} = 4000$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{4000}{80} = 50$

Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$  and

$\sum f_i|x_i - \overline{x}|$ = 1280

$\therefore$ $M.D.(\overline{x}) = \frac{1}{80}\sum_{i=1}^{5}f_i|x_i - \overline{x}|$

$= \frac{1280}{80} = 16$

Hence, the mean deviation about the mean is 16

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

$\small x_i$            $\small 5$            $\small 7$            $\small 9$            $\small 10$            $\small 12$            $\small 15$

$\small f_i$             $\small 8$            $\small 6$            $\small 2$              $\small 2$             $\small 2$              $\small 6$

 $x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 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 $\dpi{100} 13^{th}$ and $\dpi{100} 14^{th}$ observations.

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

Therefore, Median, M  $\dpi{100} = \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$  and

$\sum f_i|x_i - M|$ = 84

$\therefore$ $M.D.(M) = \frac{1}{26}\sum_{i=1}^{6}|x_i - M|$

$= \frac{84}{26} = 3.23$

Hence, the mean deviation about the median is 3.23

$\small x_i$            $\small 15$            $\small 21$            $\small 27$            $\small 30$            $\small 35$

$\small f_i$            $\small 3$                $\small 5$             $\small 6$              $\small 7$             $\small 8$

 $x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 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 $15^{th}$ and $\dpi{100} 16^{th}$ observations.

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

Therefore, Median, M  $\dpi{100} = \frac{15^{th} observation + 16^{th} observation}{2} = \frac{30 + 30}{2} = 30$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$  and

$\sum f_i|x_i - M|$ = 149.5

$\therefore$ $M.D.(M) = \frac{1}{29}\sum_{i=1}^{5}|x_i - M|$

$= \frac{149.5}{29} = 5.1$

Hence, the mean deviation about the median is 5.1

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

 Income per day in Rs $\small 0-100$ $\small 100-200$ $\small 200-300$ $\small 300-400$ $\small 400-500$ $\small 500-600$ $\small 600-700$ $\small 700-800$ Number of persons $\small 4$ $\small 8$ $\small 9$ $\small 10$ $\small 7$ $\small 5$ $\small 4$ $\small 3$

 Income per day Number of  Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 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 $\sum{f_i}$ =50 $\sum f_ix_i$ =17900 $\sum f_i|x_i - \overline{x}|$ =7896

$N = \sum_{i=1}^{8}{f_i} = 50 ; \sum_{i=1}^{8}{f_ix_i} = 17900$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{8}f_ix_i = \frac{17900}{50} = 358$

Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$  and

$\sum f_i|x_i - \overline{x}|$ = 7896

$\therefore$ $M.D.(\overline{x}) = \frac{1}{50}\sum_{i=1}^{8}f_i|x_i - \overline{x}|$

$= \frac{7896}{50} = 157.92$

Hence, the mean deviation about the mean is 157.92

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

 Height in cms $\small 95-105$ $\small 105-115$ $\small 115-125$ $\small 125-135$ $\small 135-145$ $\small 145-155$ Number of persond $\small 9$ $\small 13$ $\small 26$ $\small 30$ $\small 12$ $\small 10$

 Height in cms Number of  Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 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 $\sum{f_i}$ =100 $\sum f_ix_i$ =12530 $\sum f_i|x_i - \overline{x}|$ =1128.8

$N = \sum_{i=1}^{6}{f_i} = 100 ; \sum_{i=1}^{6}{f_ix_i} = 12530$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{6}f_ix_i = \frac{12530}{100} = 125.3$

Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$  and

$\sum f_i|x_i - \overline{x}|$ = 1128.8

$\therefore$ $M.D.(\overline{x}) = \frac{1}{100}\sum_{i=1}^{6}f_i|x_i - \overline{x}|$

$= \frac{1128.8}{100} = 11.29$

Hence, the mean deviation about the mean is 11.29

 Marks $\small 0-10$ $\small 10-20$ $\small 20-30$ $\small 30-40$ $\small 40-50$ $\small 50-60$ Number of girls $\small 6$ $\small 8$ $\small 14$ $\small 16$ $\small 4$ $\small 2$

 Marks Number of  Girls $f_i$ Cumulative Frequency c.f. Mid Points $x_i$ $|x_i - M|$ $f_i|x_i - M|$ 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 $\sum f_i|x_i - M|$ =517.1

Now, N = 50, which is even.

The class interval containing $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $\dpi{100} 25^{th}$ item is 20-30. Therefore, 20-30 is the median class.

We know,

Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$

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

Therefore, Median $\dpi{100} = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$  and

$\sum f_i|x_i - M|$ = 517.1

$\therefore$ $M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|$

$= \frac{517.1}{50} = 10.34$

Hence, the mean deviation about the median is 10.34

 Age (in years) $\small 16-20$ $\small 21-25$ $\small 26-30$ $\small 31-35$ $\small 36-40$ $\small 41-45$ $\small 46-50$ $\small 51-55$ Number $\small 5$ $\small 6$ $\small 12$ $\small 14$ $\small 26$ $\small 12$ $\small 16$ $\small 9$

[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]

 Age (in years) Number  $f_i$ Cumulative Frequency c.f. Mid Points $x_i$ $|x_i - M|$ $f_i|x_i - M|$ 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 $\sum f_i|x_i - M|$ =735

Now, N = 100, which is even.

The class interval containing $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $50^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$

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

Therefore, Median $\dpi{100} = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$

Now, we calculate the absolute values of the deviations from median, $|x_i - M|$  and

$\sum f_i|x_i - M|$ = 735

$\therefore$ $M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|$

$= \frac{735}{100} = 7.35$

Hence, the mean deviation about the median is 7.35

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

$\small 6, 7, 10, 12, 13, 4, 8, 12$

Mean ($\overline{x}$) of the given data:

$\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{6+ 7+ 10+ 12+ 13+ 4+ 8+ 12}{8} = \frac{72}{8} = 9$

The respective values of the deviations from mean, $(x_i - \overline{x})$ are

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

$\therefore$  $\sum_{i=1}^{8}(x_i - 10)^2 = 74$

$\therefore$ $\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\frac{1}{8}\sum_{i=1}^{8}(x_i - \overline{x})^2= \frac{74}{8} = 9.25$

Hence,  Mean = 9 and Variance = 9.25

First n natural numbers.

Mean ($\overline{x}$) of first n natural numbers:

$\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$

We know, Variance$\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}\left (x_i - \frac{n+1}{2} \right )^2$

We know that $(a-b)^2 = a^2 - 2ab + b^2$

$\\ \therefore n\sigma^2 = \sum_{i=1}^{n}x_i^2 + \sum_{i=1}^{n}(\frac{n+1}{2})^2 - 2\sum_{i=1}^{n}x_i\frac{n+1}{2} \\ = \frac{n(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4}\times n - 2.\frac{(n+1)}{2}.\frac{n(n+1)}{2}$

$\\ \implies \sigma^2 = \frac{(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4} - \frac{(n+1)^2}{2}$

$\\ = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \\ = (n+1)\left [\frac{4n+2 - 3n -3}{12} \right ] \\ = (n+1).\frac{(n-1)}{12} \\ = \frac{n^2-1}{12}$

Hence, Mean = $\frac{n+1}{2}$ and Variance = $\frac{n^2-1}{12}$

First 10 multiples of 3

First 10 multiples of 3 are:

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

Mean ($\overline{x}$) of the above values:

$\\ \overline{x} = \frac{1}{10}\sum_{i=1}^{10}x_i = \frac{3+ 6+ 9+ 12+ 15+ 18+ 21+ 24+ 27+ 30}{10} \\ = 3.\frac{\frac{10(10+1)}{2}}{10} = 16.5$

The respective values of the deviations from mean, $(x_i - \overline{x})$ are

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

$\therefore$  $\sum_{i=1}^{10}(x_i - 16.5)^2 = 742.5$

$\therefore$ $\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\frac{1}{10}\sum_{i=1}^{10}(x_i - \overline{x})^2= \frac{742.5}{10} = 74.25$

Hence,  Mean = 16.5 and Variance = 74.25

 $\small x_i$ 6 10 14 18 24 28 30 $\small f_i$ 2 4 7 12 8 4 3

 $x_i$ $f_i$ $f_ix_i$ $(x_i - \overline{x})$ $(x_i - \overline{x})^2$ $f_i(x_i - \overline{x})^2$ 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 13 169 363 $\sum{f_i}$ = 40 $\sum f_ix_i$ = 760 $\sum f_i(x_i - \overline{x})^2$ =1736

$N = \sum_{i=1}^{7}{f_i} = 40 ; \sum_{i=1}^{7}{f_ix_i} = 760$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{760}{40} = 19$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\implies \sigma^2 = \frac{1736}{40} = 43.4$

Hence, Mean = 19 and Variance = 43.4

Question:â€‹â€‹â€‹â€‹â€‹â€‹â€‹5. Find the mean and variance for each of the data.

 $\small x_i$ 92 93 97 98 102 104 109 $\small f_i$ 3 2 3 2 6 3 3

 $x_i$ $f_i$ $f_ix_i$ $(x_i - \overline{x})$ $(x_i - \overline{x})^2$ $f_i(x_i - \overline{x})^2$ 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 $\sum{f_i}$ = 22 $\sum f_ix_i$ = 2200 $\sum f_i(x_i - \overline{x})^2$ =640

$N = \sum_{i=1}^{7}{f_i} = 22 ; \sum_{i=1}^{7}{f_ix_i} = 2200$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{2200}{22} = 100$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\implies \sigma^2 = \frac{640}{22} = 29.09$

Hence, Mean = 100 and Variance = 29.09

Question:â€‹â€‹â€‹â€‹â€‹â€‹â€‹6     Find the mean and standard deviation using short-cut method.

 $\small x_i$ 60 61 62 63 64 65 66 67 68 $\small f_i$ 2 1 12 29 25 12 10 4 5

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

 $x_i$ $f_i$ $\dpi{100} y_i = \frac{x_i-A}{h}$ $y_i^2$ $f_iy_i$ $f_iy_i^2$ 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 $\sum{f_i}$ =100 $\sum f_iy_i$ = 0 $\sum f_iy_i ^2$ =286

$N = \sum_{i=1}^{9}{f_i} = 100 ; \sum_{i=1}^{9}{f_iy_i} = 0$

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =64 + \frac{0}{100} = 64$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(286) - (0)^2 \right ] \\ = \frac{28600}{10000} = 2.86$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{2.86} = 1.691$

Hence, Mean = 64 and Standard Deviation = 1.691

Question:â€‹â€‹â€‹â€‹â€‹â€‹â€‹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 $f_i$ Mid point  $x_i$ $f_ix_i$ $(x_i - \overline{x})$ $(x_i - \overline{x})^2$ $f_i(x_i - \overline{x})^2$ 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 $\sum{f_i}$ = N = 30 $\sum f_ix_i$ = 3210 $\sum f_i(x_i - \overline{x})^2$ =68280

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{3210}{30} = 107$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\implies \sigma^2 = \frac{68280}{30} = 2276$

Hence, Mean = 107 and Variance = 2276

Question:â€‹â€‹â€‹â€‹â€‹â€‹â€‹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 $f_i$ Mid-point  $x_i$ $f_ix_i$ $(x_i - \overline{x})$ $(x_i - \overline{x})^2$ $f_i(x_i - \overline{x})^2$ 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 $\sum{f_i}$ = N = 50 $\sum f_ix_i$ = 1350 $\sum f_i(x_i - \overline{x})^2$ =6600

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{1350}{50} = 27$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\implies \sigma^2 = \frac{6600}{50} = 132$

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 $f_i$ Midpoint $x_i$ $\dpi{100} y_i = \frac{x_i-A}{h}$ $y_i^2$ $f_iy_i$ $f_iy_i^2$ 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 $\sum{f_i}$ =N = 60 $\sum f_iy_i$ = 6 $\sum f_iy_i ^2$ =254

Mean,

$\overline{y} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =92.5 + \frac{6}{60}\times5 = 93$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(60)^2}\left [60(254) - (6)^2 \right ] \\ = \frac{1}{(60)^2}\left [15240 - 36 \right ] \\ = \frac{15204}{144} = 105.583$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{105.583} = 10.275$

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 $32.5-36.5,36.5-40.4,40.5-44.5,44.5-48.5,48.5-52.5$  and then proceed.]

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

 Diameters No. of circles $f_i$ Midpoint $x_i$ $\dpi{100} y_i = \frac{x_i-A}{h}$ $y_i^2$ $f_iy_i$ $f_iy_i^2$ 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 $\sum{f_i}$ =N = 100 $\sum f_iy_i$ = 25 $\sum f_iy_i ^2$ =199

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =42.5 + \frac{25}{100}\times4 = 43.5$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(199) - (25)^2 \right ]\times4^2 \\ = \frac{1}{625}\left [19900 - 625 \right ] \\ = \frac{19275}{625} = 30.84$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{30.84} = 5.553$

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:â€‹â€‹â€‹â€‹â€‹â€‹â€‹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 $f_i$ Midpoint $x_i$ $\dpi{100} y_i = \frac{x_i-A}{h}$ $\dpi{80} = \frac{x_i-45}{10}$ $y_i^2$ $f_iy_i$ $f_iy_i^2$ 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 $\sum{f_i}$ =N = 150 $\sum f_iy_i$ = -6 $\sum f_iy_i ^2$ =342

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(342) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [51264 \right ] \\ =227.84$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{227.84} = 15.09$

Coefficient of variation = $\frac{\sigma}{\overline x}\times100$

C.V.(A) = $\frac{15.09}{44.6}\times100 = 33.83$

Similarly,

For Group B

 Marks Group A $f_i$ Midpoint $x_i$ $\dpi{100} y_i = \frac{x_i-A}{h}$ $\dpi{80} = \frac{x_i-45}{10}$ $y_i^2$ $f_iy_i$ $f_iy_i^2$ 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 72 $\sum{f_i}$ =N = 150 $\sum f_iy_i$ = -6 $\sum f_iy_i ^2$ =375

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(375) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [56214 \right ] \\ =249.84$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{249.84} = 15.80$

Coefficient of variation = $\frac{\sigma}{\overline x}\times100$

C.V.(B) = $\frac{15.80}{44.6}\times100 = 35.42$

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($x_i$) Y($y_i$) $x_i^2$ $y_i^2$ 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 , $\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{510}{10} = 51$

Variance, $\sigma^2 = \frac{1}{n^2}\left [n\sum x_i^2 - (\sum x_i)^2 \right ]$

$\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(26360) - (510)^2 \right ] \\ = \frac{1}{100}.\left [263600 - 260100 \right ] \\ \\ = 35$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{35} = 5.91$

C.V.(X) = $\frac{\sigma}{\overline x}\times100 = \frac{5.91}{51}\times100 = 11.58$

Similarly, For Y,

Mean , $\overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i = \frac{1050}{10} = 105$

Variance, $\sigma^2 = \frac{1}{n^2}\left [n\sum y_i^2 - (\sum y_i)^2 \right ]$

$\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(110290) - (1050)^2 \right ] \\ = \frac{1}{100}.\left [1102900 - 1102500 \right ] \\ \\ = 4$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{4} = 2$

C.V.(Y) = $\frac{\sigma}{\overline y}\times100 = \frac{2}{105}\times100 = 1.904$

Since C.V.(Y) < C.V.(X)

Hence Y is more stable.

 Firm A Firm B No. of wage earners $586$ $648$ Mean of monthly wages $Rs\hspace {1mm} 5253$ $Rs\hspace {1mm} 5253$ Variance of the distribution of wages $100$ $121$

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 $586$ $648$ Mean of monthly wages $Rs\hspace {1mm} 5253$ $Rs\hspace {1mm} 5253$ Variance of the distribution of wages $100$ $121$

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

Standard Deviation = $\sigma_A = \sqrt{Variance}= \sqrt{100} = 10$
Standard Deviation = $\sigma_B = \sqrt{Variance}= \sqrt{121} = 11$
Since $\sigma_B>\sigma_A$, firm B has greater variability in individual wages.