NCERT Solutions for Class 11 Maths Chapter 15 Statistics

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NCERT Solutions for Class 11 Maths Chapter 15 Statistics

Edited By Ramraj Saini | Updated on Sep 25, 2023 10:18 PM IST

Statistics Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 15 Statistics are provided here. You have learnt the basic statistics like mean, mode and median known as "measures of the central tendency" in the previous Class NCERT syllabus. In this NCERT book chapter, you will learn important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range and many more. In class 11 maths chapter 15 question answer, you will get questions related to all the above topics. These NCERT solutions are prepared by expert team at Careers360 keeping in my the latest syllabus of CBSE 2023-24, easy and simple language, step by step comprehensive coverage of the concepts.

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Statistics Class 11 Questions And Answers
Statistics Class 11 Questions And Answers PDF Free Download
Statistics Class 11 Solutions - Important Formulae
Statistics Class 11 NCERT Solutions (Intext Questions and Exercise)
Summary And Formulae Of Statistics Maths Chapter 15 class 11
NCERT Solutions For Class 11 Mathematics
Key Features Of Statistics Class 11 NCERT Solutions
NCERT Solutions For Class 11 - Subject Wise
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 11 Maths Chapter 15 Statistics

This statistics class 11 solutions 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. Class 11 maths chapter 15 NCERT solutions 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. Interested students can find all NCERT solutions for class 11 at one place which will help to understand the concepts in an easy way.

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Statistics Class 11 Questions And Answers PDF Free Download

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Statistics Class 11 Solutions - Important Formulae

Measure of Dispersion:

Dispersion measures the degree of variation in the values of a variable. It quantifies how scattered observations are around the central value in a distribution.

Range:

Range is the simplest measure of dispersion. It is defined as the difference between the largest and smallest observations in a distribution.

Range of distribution = Largest observation – Smallest observation.

Mean Deviation:

Mean Deviation for Ungrouped Data:

For n observations x1, x2, x3, ..., xn, the mean deviation about their mean x¯ is calculated as:
- MD(\bar x) = (Σ |xᵢ - x¯|) / n
The mean deviation about its median M is calculated as:
- MD(M) = (Σ |xᵢ - M|) / n

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Mean Deviation for Discrete Frequency Distribution:

For a discrete frequency distribution with values xᵢ, frequencies fᵢ, mean x¯, and total frequency N, the mean deviation is calculated as:
- MD(\bar x) = (Σ fᵢ |xᵢ - x¯|) / (Σ fᵢ) = (Σ fᵢ |xᵢ - x¯|) / N

Variance:

Variance is the average of the squared deviations from the mean x¯. If x_₁, x_₂, …, xₙ are n observations with mean x¯, the variance denoted by σ² is calculated as:
- σ^² = (Σ(xᵢ - x¯)^²) / n

Standard Deviation:

Standard deviation, denoted as σ, is the square root of the variance σ^². If σ^² is the variance, then the standard deviation is given by:
- σ = √(σ^²)
For a discrete frequency distribution with values xᵢ, frequencies fᵢ, mean x¯, and total frequency N, the standard deviation is calculated as:
- σ = √((Σ fᵢ(xᵢ - x¯)^²) / N)

Coefficient of Variation:

The coefficient of variation (CV) is used to compare two or more frequency distributions. It is defined as:
- Coefficient of Variation = (Standard Deviation / Mean) × 100
- CV = (σ / x¯) × 100

Free download NCERT Solutions for Class 11 Maths Chapter 15 Statistics for CBSE Exam.

Statistics Class 11 NCERT Solutions (Intext Questions and Exercise)

Class 11 maths chapter 15 question answer - Exercise: 15.1

Question:1 . Find the mean deviation about the mean for the data.

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

Answer:

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.

Question:2. Find the mean deviation about the mean for the data.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

$x_i$	$f_i$	$f_ix_i$	$\|x_i - \overline{x}\|$	$f_i\|x_i - \overline{x}\|$
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.

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

Answer:

$x_i$	$f_i$	$f_ix_i$	$\|x_i - \overline{x}\|$	$f_i\|x_i - \overline{x}\|$
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.

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

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

Answer:

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

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

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

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

Question:8 Find the mean deviation about the median.

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

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

Answer:

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

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

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

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	$\small 4$	$\small 8$	$\small 9$	$\small 10$	$\small 7$	$\small 5$	$\small 4$	$\small 3$

Answer:

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	$\small 9$	$\small 13$	$\small 26$	$\small 30$	$\small 12$	$\small 10$

Answer:

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

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

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

Answer:

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

We know,

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

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

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

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

Question:12 Calculate the mean deviation about median age for the age distribution of $\small 100$ persons given below:

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]

Answer:

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

We know,

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

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

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

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

Class 11 maths chapter 15 ncert solutions - Exercise: 15.2

Question:1. Find the mean and variance for each of the data.

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

Answer:

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

Question:2. Find the mean and variance for each of the data.

First n natural numbers.

Answer:

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 = $\frac{n^2-1}{12}$

Question:3. Find the mean and variance for each of the data

First 10 multiples of 3

Answer:

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

Question:4. Find the mean and variance for each of the data.

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

Answer:

$x_i$	$f_i$	$f_ix_i$	$(x_i - \overline{x})$	$(x_i - \overline{x})^2$	$f_i(x_i - \overline{x})^2$
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: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

Answer:

$x_i$	$f_i$	$f_ix_i$	$(x_i - \overline{x})$	$(x_i - \overline{x})^2$	$f_i(x_i - \overline{x})^2$
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: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

Answer:

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

$x_i$	$f_i$	$y_i = \frac{x_i-A}{h}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
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:âââââââ7. 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

Answer:

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:âââââââ8. 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

Answer:

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

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

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

Answer:

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

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

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.]

Answer:

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

Statistics class 11 NCERT solutions - 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

Answer:

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 $x_i$	$\dpi{100} y_i = \frac{x_i-A}{h}$ $= \frac{x_i-45}{10}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
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 $x_i$	$\dpi{100} y_i = \frac{x_i-A}{h}$ $= \frac{x_i-45}{10}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
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.

Question:2 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

Answer:

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.

Question:3(i) 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	$Rs\hspace {1mm} 5253$	$Rs\hspace {1mm} 5253$
Variance of the distribution of wages	$100$	$121$

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

Answer:

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.

Question:3(ii) 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	$Rs\hspace {1mm} 5253$	$Rs\hspace {1mm} 5253$
Variance of the distribution of wages	$100$	$121$

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

Answer:

Given, Variance of firm A = 100

Standard Deviation = $\sigma_A = \sqrt{Variance}= \sqrt{100} = 10$

Again, Variance of firm B = 121

Standard Deviation = $\sigma_B = \sqrt{Variance}= \sqrt{121} = 11$

Since $\sigma_B>\sigma_A$ , firm B has greater variability in individual wages.

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

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 $1.25$ goals. Find which team may be considered more consistent?

Answer:

No. of goals scored $x_i$	Frequency $f_i$	$x_i^2$	$f_ix_i$	$f_ix_i^2$
0	1	0	0	0
1	9	1	9	9
2	7	4	14	28
3	5	9	15	45
4	3	16	12	48
	$\sum{f_i}$ =N = 25		$\sum f_ix_i$ = 50	$\sum f_ix_i ^2$ =130

For Team A,

Mean,

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i =\frac{50}{25}= 2$

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

$\\ \implies \sigma^2 = \frac{1}{(25)^2}\left [25(130) - (50)^2 \right ] \\ \\ = \frac{750}{625} =1.2$

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

$\therefore \sigma = \sqrt{1.2} = 1.09$

C.V.(A) = $\frac{\sigma}{\overline x}\times100 = \frac{1.09}{2}\times100 = 54.5$

For Team B,

Mean = 2

Standard deviation, $\sigma$ = 1.25

C.V.(B) = $\frac{\sigma}{\overline x}\times100 = \frac{1.25}{2}\times100 = 62.5$

Since C.V. of firm B is more than C.V. of A.

Therefore, Team A is more consistent.

Question: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:

$\sum_{i=1}^{50}x_i=212, \sum _{i=1}^{50}x_i^2=902.8,\sum _{i=1}^{50}y_i=261,\sum _{i=1}^{50}y_i^2=1457.6$
Which is more varying, the length or weight?

Answer:

For lenght x,

Mean, $\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i =\frac{212}{50}= 4.24$

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

$\\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(902.8) - (212)^2 \right ] \\ \\ = \frac{196}{2500} =0.0784$

We know, Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{0.0784} = 0.28$

C.V.(x) = $\frac{\sigma}{\overline x}\times100 = \frac{0.28}{4.24}\times100 = 6.603$

For weight y,

Mean,

Mean, $\overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i =\frac{261}{50}= 5.22$

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

$\\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(1457.6) - (261)^2 \right ] \\ \\ = \frac{4759}{2500} =1.9036$

We know, Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{1.9036} = 1.37$

C.V.(y) = $\frac{\sigma}{\overline y}\times100 = \frac{1.37}{5.22}\times100 = 26.24$

Since C.V.(y) > C.V.(x)

Therefore, weight is more varying.

Statistics maths chapter 15 class 11 - Miscellaneous Exercise

Question: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.

Answer:

Given,

The mean and variance of 8 observations are 9 and 9.25, respectively

Let the remaining two observations be x and y,

Observations: 6, 7, 10, 12, 12, 13, x, y.

∴ Mean, $\overline X = \frac{6+ 7+ 10+ 12+ 12+ 13+ x+ y}{8} = 9$

60 + x + y = 72

x + y = 12 -(i)

Now, Variance

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

$\implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2 -18(x+y)+ 2.9^2 \right ]$

$\implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2+ -18(12)+ 2.9^2 \right ]$ (Using (i))

$\implies 9.25 = \frac{1}{8}\left[48+x^2+ y^2 -216+ 162 \right ] = \frac{1}{8}\left[x^2+ y^2 - 6 \right ]$

$\implies x^2+ y^2 = 80$ -(ii)

Squaring (i), we get

$x^2+ y^2 +2xy= 144$ (iii)

(iii) - (ii) :

2xy = 64 (iv)

Now, (ii) - (iv):

$\\ x^2+ y^2 -2xy= 80-64 \\ \implies (x-y)^2 = 16 \\ \implies x-y = \pm 4$ -(v)

Hence, From (i) and (v):

x – y = 4 $\implies$ x = 8 and y = 4

x – y = -4 $\implies$ x = 4 and y = 8

Therefore, The remaining observations are 4 and 8. (in no order)

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

Answer:

Given,

The mean and variance of 7 observations are 8 and 16, respectively

Let the remaining two observations be x and y,

Observations: 2, 4, 10, 12, 14, x, y

∴ Mean, $\overline X = \frac{2+ 4+ 10+ 12+ 14+ x+ y}{7} = 8$

42 + x + y = 56

x + y = 14 -(i)

Now, Variance

$= \frac{1}{n}\sum_{i=1}^8(x_i - \overline x)^2 = 16$

$\implies 16 = \frac{1}{7}\left[(-6)^2+ (-4)^2+ 2^2+ 4^2+ 6^2+ x^2+ y^2 -16(x+y)+ 2.8^2 \right ]$

$\implies 16 = \frac{1}{7}\left[36+16+4+16+36+ x^2+ y^2 -16(14)+ 2(64) \right ]$ (Using (i))

$\implies 16 = \frac{1}{7}\left[108+x^2+ y^2 -96 \right ] = \frac{1}{7}\left[x^2+ y^2 + 12\right ]$

$\implies x^2+ y^2 = 112- 12 =100$ -(ii)

Squaring (i), we get

$x^2+ y^2 +2xy= 196$ (iii)

(iii) - (ii) :

2xy = 96 (iv)

Now, (ii) - (iv):

$\\ x^2+ y^2 -2xy= 100-96 \\ \implies (x-y)^2 = 4 \\ \implies x-y = \pm 2$ -(v)

Hence, From (i) and (v):

x – y = 2 $\implies$ x = 8 and y = 6

x – y = -2 $\implies$ x = 6 and y = 8

Therefore, The remaining observations are 6 and 8. (in no order)

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

Answer:

Given,

Mean = 8 and Standard deviation = 4

Let the observations be $x_1, x_2, x_3, x_4, x_5\ and\ x_6$

Mean, $\overline x = \frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} = 8$

Now, Let $y_i$ be the the resulting observations if each observation is multiplied by 3:

$\\ \overline y_i = 3\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{3}$

New mean, $\overline y = \frac{y_1+ y_2+ y_3+ y_4+ y_5+ y_6}{6}$

$= 3\left [\frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} \right] =3\times 8$

= 24

We know that,

Standard Deviation = $\sigma = \sqrt{Variance}$

$\dpi{100} =\sqrt{ \frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2}$

$\dpi{100}\\ \implies 4^2=\frac{1}{6}\sum_{i=1}^6(x_i - \overline x)^2 \\ \implies \sum_{i=1}^6(x_i - \overline x)^2 = 6\times16 = 96$ -(i)

Now, Substituting the values of $x_i\ and\ \overline x$ in (i):

$\dpi{100} \\ \implies \sum_{i=1}^6(\frac{y_i}{3} - \frac{\overline y}{3})^2 = 96 \\ \implies \sum_{i=1}^6(y_i - \overline y)^2 = 96\times9 =864$

Hence, the variance of the new observations = $\frac{1}{6}\times864 = 144$

Therefore, Standard Deviation = $\sigma = \sqrt{Variance}$ = $\sqrt{144}$ = 12

Question:4 Given that $\small \bar {x}$ is the mean and $\small \sigma^2$ is the variance of $\small n$ observations .Prove that the mean and variance of the observations $\small ax_1,ax_2,ax_3,....,ax_n$ , are $\small a\bar{x}$ and $\small a^2 \sigma^2$ respectively, $\small (a \neq 0)$ .

Answer:

Given, Mean = $\small \bar {x}$ and variance = $\small \sigma^2$

Now, Let $y_i$ be the the resulting observations if each observation is multiplied by a:

$\\ \overline y_i = a\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{a}$

$\overline y =\frac{1}{n}\sum_{i=1}^ny_i = \frac{1}{n}\sum_{i=1}^nax_i$

$\overline y = a\left [\frac{1}{n}\sum_{i=1}^nx_i \right] = a\overline x$

Hence the mean of the new observations $\small ax_1,ax_2,ax_3,....,ax_n$ is $\small a\bar{x}$

We know,

$\dpi{100} \sigma^2=\frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2$

Now, Substituting the values of $x_i\ and\ \overline x$ :

$\\ \implies \sigma^2= \frac{1}{n}\sum_{i=1}^n(\frac{y_i}{a} - \frac{\overline y}{a})^2 \\ \implies a^2\sigma^2= \frac{1}{n}\sum_{i=1}^n(y_i - \overline y)^2$

Hence the variance of the new observations $\small ax_1,ax_2,ax_3,....,ax_n$ is $a^2\sigma^2$

Hence proved.

Question:5(i) The mean and standard deviation of $\small 20$ observations are found to be $\small 10$ and $\small 2$ , respectively. On rechecking, it was found that an observation $\small 8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

If wrong item is omitted.

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

$\overline x =\frac{1}{n}\sum_{i=1}^nx_i$

$\implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i$ $\implies \sum_{i=1}^{20}x_i = 200$

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 = 192

Therefore, Correct Mean = (Correct Sum)/19

$=\frac{192}{19}$

= 10.1

Now, Standard Deviation,

$\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}$

$\\ \implies 2^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 4 + (\overline x)^2 \\ \implies \frac{1}{20}\sum_{i=1}^nx_i ^2 = 4 + 100 = 104 \\ \implies \sum_{i=1}^nx_i ^2 = 2080$ ,which is the incorrect sum.

Thus, New sum = Old sum - (8x8)

= 2080 – 64

= 2016

Hence, Correct Standard Deviation =

$\sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2016}{19} - (10.1)^2}$

$\dpi{100} = \sqrt{106.1 - 102.01} = \sqrt{4.09}$

=2.02

Question:5(ii) The mean and standard deviation of $\small 20$ observations are found to be $\small 10$ and $\small 2$ , respectively. On rechecking, it was found that an observation $\small 8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

If it is replaced by $\small 12.$

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

$\overline x =\frac{1}{n}\sum_{i=1}^nx_i$

$\implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i$ $\implies \sum_{i=1}^{20}x_i = 200$

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 + 12 = 204

Therefore, Correct Mean = (Correct Sum)/20

$\dpi{100} =\frac{204}{20}$

= 10.2

Now, Standard Deviation,

$\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}$

Thus, New sum = Old sum - (8x8) + (12x12)

= 2080 – 64 + 144

= 2160

Hence, Correct Standard Deviation =

$\sigma' =\sqrt{\frac{1}{n}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2160}{20} - (10.2)^2}$

$= \sqrt{108 - 104.04} = \sqrt{3.96}$

= 1.98

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

Subject	Mathematics	Physics	Chemistry
Mean	$\small 42$	$\small 32$	$\small 40.9$
Standard deviation	$\small 12$	$\small 15$	$\small 20$

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

Answer:

Given,

Standard deviation of physics = 15

Standard deviation of chemistry = 20

Standard deviation of mathematics = 12

We know ,

C.V. $= \frac{Standard\ Deviation}{Mean}\times 100$

The subject with greater C.V will be more variable than others.

$C.V. _P= \frac{15}{32}\times 100 = 46.87$

$C.V. _C= \frac{20}{40.9}\times 100 = 48.89$

$C.V. _M= \frac{12}{42}\times 100 = 28.57$

Hence, Mathematics has lowest variability and Chemistry has highest variability.

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

Answer:

Given,

Initial Number of observations, n = 100

$\overline x =\frac{1}{n}\sum_{i=1}^nx_i$

$\dpi{100} \implies 20 =\frac{1}{100}\sum_{i=1}^{100}x_i$ $\implies \sum_{i=1}^{100}x_i = 2000$

Thus, incorrect sum = 2000

Hence, New sum of observations = 2000 - 21-21-18 = 1940

New number of observation, n' = 100-3 =97

Therefore, New Mean = New Sum)/100

$=\frac{1940}{97}$

= 20

Now, Standard Deviation,

$\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}$

$\\ \implies 3^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 9 + (\overline x)^2 \\ \implies \frac{1}{100}\sum_{i=1}^nx_i ^2 = 9 + 400 = 409 \\ \implies \sum_{i=1}^nx_i ^2 = 40900$ ,which is the incorrect sum.

Thus, New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18)

= 40900 - 441 - 441 - 324

= 39694

Hence, Correct Standard Deviation =

$\sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{39694}{97} - (20)^2}$

$= \sqrt{108 - 104.04} = \sqrt{3.96}$

= 3.036

Summary And Formulae Of Statistics Maths Chapter 15 class 11

Measures of dispersion- Range, mean deviation, Quartile deviation, standard deviation, variance are measures of dispersion.
Range- The difference between the maximum and the minimum values of each series is called the ‘Range’ of the data. Range = Maximum Value – Minimum Value.
Mean- The arithmetic average of a range of values or quantities, found by adding all values and dividing by the number of values. For example, the mean of 4, 10, and 1 is (4+10+1)/3=15/3=5.

$\bar{X}=\frac{\Sigma X}{N}$
X represents values or scores,
N represents the number of values or scores.

Median- If the total number of values(n) is an odd number, then the formula is-

$Median= (\frac{n+1}{2})^{th}\:term$

Median- If the total number of values(n) is an even number, then the formula is-

$Median= \frac{(\frac{n}{2})^{th}\:term+(\frac{n+1}{2})^{th}\:term}{2}$

Mode- It is the most frequently occurring value or number.
Mean deviation for ungrouped data-

$M.D.(\bar{x})=\frac{\Sigma \:|x_{i}-\bar{x}|}{n}, \:\:\:\:\:\:\:\:\: M.D.(M)=\frac{\Sigma \:|x_{i}-M|}{n}$

Mean deviation for grouped data-

$M.D.(\bar{x})=\frac{\Sigma f_i\:|x_{i}-\bar{x}|}{N}, \:\:\:\:\:\:\:\:\: M.D.(M)=\frac{f_i\Sigma \:|x_{i}-M|}{N}, \: where \:N=\Sigma f_i$

interested students can practice class 11 maths ch 15 question answer using the following exercises.

NCERT Solutions For Class 11 Mathematics

chapter-1	Sets
chapter-2	Relations and Functions
chapter-3	Trigonometric Functions
chapter-4	Principle of Mathematical Induction
chapter-5	Complex Numbers and Quadratic equations
chapter-6	Linear Inequalities
chapter-7	Permutation and Combinations
chapter-8	Binomial Theorem
chapter-9	Sequences and Series
chapter-10	Straight Lines
chapter-11	Conic Section
chapter-12	Three Dimensional Geometry
chapter-13	Limits and Derivatives
chapter-14	Mathematical Reasoning
chapter-15	Statistics
chapter-16	Probability

Key Features Of Statistics Class 11 NCERT Solutions

Easy to Understand: The ch 15 maths class 11 solutions are written in simple language and are easy to understand, which makes them accessible for students who may not have a strong background in statistics.

Comprehensive Coverage: The statistics NCERT solutions cover all the important concepts and topics in Statistics Class 11 NCERT curriculum. This ensures that students are well-prepared and have a thorough understanding of the subject.

Structured Format: The ch 15 maths class 11 solutions are presented in a structured format that helps students to organize their thoughts and approach problems in a systematic manner.

NCERT Solutions For Class 11 - Subject Wise

NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

NCERT solutions for Class 11 physics

This chapter seems very easy but it very useful in the field of science, research reliable predictions, or forecasts for future use, decision making, etc. There are 7 questions are given in the miscellaneous exercise. Try to solve them also to get command in this chapter. In NCERT solutions for class 11 maths chapter 15 statistics, you will get solutions to miscellaneous exercise too.

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. Explain the term standard deviation, which is discussed in NCERT solutions for class 11 maths chapter 15.

The concept of standard deviation involves measuring the amount of variation or deviation present in a given set of values, and the range is determined by considering the level of standard deviation. For a clearer understanding of this term, students are encouraged to download the PDF solutions provided by Careers360, which are available in both chapter-wise and exercise-wise formats to cater to their specific needs.

2. What are the topics discussed in Chapter 15 of NCERT Solutions for Class 11 Maths?

The topics discussed in class 11 maths statistics are
1. Introduction
2. Methods of Dispersion
3. Range
4. Mean Deviation
5. Variance and Standard Deviation
6. Analysis of Frequency Distributions

3. Are the NCERT Solutions for statistics class 11 maths helpful for the students?

The NCERT Solutions for class 11 chapter 15 maths offer precise and straightforward explanations that aid students in achieving good grades in their exams. The solutions' methodical approach to problem-solving provides students with a clear understanding of the marks allocation as per the latest CBSE Syllabus 2023. By using these class 11th statistics solutions , students can identify their weak areas and work towards improving them, resulting in better academic performance.

4. Where can I find the complete solutions of class 11 statistics solutions ?

Students can find a detailed NCERT solutions for class 11 maths by clicking on the link. they can practice these solutions to get confidence in the concepts and in-depth understanding that ultimately lead to high score in the exam.

Also Read

NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

NCERT Class 11 Chemistry Chapter 3 Notes - Download PDF

NCERT Class 11 Biology Chapter 19 Notes Excretory Products And Their Elimination- Download PDF Notes

NCERT Class 11 Biology Chapter 22 Notes Chemical Coordination And Integration- Download PDF Notes

Some basic concepts of Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 1 Notes - Download PDF

NCERT Class 11 Biology Chapter 21 Notes Neural Control And Coordination- Download PDF Notes

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NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

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NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 11 - Limits and Derivatives

NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11 - Conic Section

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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NCERT Solutions for Class 11 Maths Chapter 15 Statistics

Statistics Class 11 Questions And Answers

Statistics Class 11 Questions And Answers PDF Free Download

Statistics Class 11 Solutions - Important Formulae

Statistics Class 11 NCERT Solutions (Intext Questions and Exercise)

Summary And Formulae Of Statistics Maths Chapter 15 class 11

NCERT Solutions For Class 11 Mathematics

Key Features Of Statistics Class 11 NCERT Solutions

NCERT Solutions For Class 11 - Subject Wise

NCERT Books and NCERT Syllabus

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