Chapter 13 Statistics

NCERT Solutions Class 11 Maths Chapter 13 – Statistics

NCERT Solutions for Class 11 Maths Chapter 13 (Statistics) helps students to move from basic data handling to the study of the Dispersion. While class 10 was about central tendency (Mean, Median, Mode) this chapter is about how spread out the data is from these central values. The knowledge of dispersion is important in risk assessment in finance, quality control in manufacturing and analysis of experimental data in science.

The NCERT Solutions for Class 11 Maths Chapter 13 of ALLEN are well prepared by expert faculty to simplify the calculation heavy nature of statistics by following organised steps and clear formulas. Solutions highlight the accuracy needed to find Variance and Standard Deviation, helping students steer clear of common arithmetic blunders.

Statistics is a regular feature in competitive exams and to ace the measures of dispersion is very important as it can fetch you some easy marks if your concepts are clear. These solutions offer a rigorous framework for handling ungrouped and grouped data. They also provide shortcut methods for computing mean and variance.

1.0Class 11 Maths Chapter 13: Key Concepts

This chapter focuses on quantifying the variability or spread in a set of data. Key lessons include:

Measures of Dispersion: Understanding that central tendency is not enough to describe data; we need to know the "scatter."
Mean Deviation (MD): The arithmetic mean of the absolute deviations of the values from a central value (Mean or Median).
- $M . D . (\overset{x}{ˉ}) = \frac{\sum ∣ x _{i} - x ˉ ∣}{n}$
Variance ( $σ^{2}$ ): The average of the squared deviations from the mean.
Standard Deviation ( $σ$ ): The positive square root of the variance, providing a measure of spread in the same units as the data.
- For grouped data: $σ = \frac{\sum f _{i} ( x _{i} - x ˉ ) ^{2}}{N}$
Shortcut Method for Variance: Learning the "Assumed Mean" method to simplify calculations involving large numbers.
Analysis of Frequency Distributions: Using the Coefficient of Variation (C.V.) to compare the stability or consistency of two different data sets.
- $C . V . = \frac{σ}{x ˉ} \times 100$

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 13

EXERCISE - 13.1

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

4, 7, 8, 9, 10, 12, 13, 17 Sol. The given data is 4, 7, 8, 9, 10, 12, 13, 17 Mean of the data,

$\overset{x}{ˉ} = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8} = \frac{80}{8} = 10$

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

$i.e. x_{i} - \overset{x}{ˉ} are - 6, - 3, - 2, - 1, 0, 2, 3, 7$

The absolute values of the deviations, i.e. $∣ x_{i} - \overline{x} ∣$ , are 6, 3, 2, 1, 0, 2, 3, 7 The required mean deviation about the mean is

$M.D. (\overset{x}{ˉ}) = \frac{\sum _{i = 1}^{8} ∣ x _{i} - x ˉ ∣}{8} = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8} = \frac{24}{8} = 3$

38, 70, 48, 40, 42, 55, 63, 46, 54, 44 Sol. The given data is

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

Mean of the given data,

$\overset{x}{ˉ} = \frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10} = \frac{500}{10} = 50$

The deviations of the respective observations from the mean $\overline{x}$ , i.e. $x_{i} - \overline{x}$ are -12, 20, -2, -10, -8, 5, 13, - 4, 4, - 6 The absolute values of the deviations, i.e. $∣ x_{i} - \overline{x} ∣$ , are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6 The required mean deviation about the mean is

$M.D. (\overline{x}) = \frac{\sum _{i = 1}^{10} ∣ x _{i} - x ∣}{10} = \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10} = \frac{84}{10} = 8.4$

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

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 Sol. The given data is

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

Here, the numbers of observations are 12, which is even. Arranging the data in ascending order, we obtain 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 Median,

$M = \frac{( \frac{12}{2} ) ^{th} observation + ( \frac{12}{2} + 1 ) ^{th} observation}{2} = \frac{6 ^{th} observation + 7 ^{th} observation}{2} = \frac{13 + 14}{2} = \frac{27}{2} = 13.5$

The deviations of the respective observations from the median, i.e. $x_{i} - M$ are

$- 3.5, - 2.5, - 2.5, - 1.5, - 0.5, - 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5$

The absolute values of the deviations, $∣ x_{i} - M ∣$ , 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 required mean deviation about the median is

$= = M . D . (M) = \frac{\sum _{i = 1}^{12} ∣ x _{i} - M ∣}{12} \frac{3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5}{12} \frac{28}{12} = 2.33$

36, 72, 46, 42, 60, 45, 53, 46, 51, 49 Sol. The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 Median,

$M = \frac{( \frac{10}{2} ) ^{th} observation + ( \frac{10}{2} + 1 ) ^{th} observation}{2} = \frac{5 ^{th} observation + 6 ^{th} observation}{2} = \frac{46 + 49}{2} = \frac{95}{2} = 47.5$

The deviations of the respective observations from the median, i.e. $x_{i} - M$ , are

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

The absolute values of the deviations, $∣ x_{i} - M ∣$ , are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5 Thus, the required mean deviation about the median is

$= = M.D. (M) = \frac{\sum _{i = 1}^{10} ∣ x _{i} - M ∣}{10} \frac{11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5}{10} \frac{70}{10} = 7$

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

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

Sol.

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$∥ x_{i} - \overline{x} ∥$	$f_{i} ∥ x_{i} - \overline{x} ∥$
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
	25	350		158

$N = i = 1 \sum 5 f_{i} = 25, i = 1 \sum 5 f_{i} x_{i} = 350 ∴ \overline{x} = \frac{1}{N} i = 1 \sum 5 f_{i} x_{i} = \frac{1}{25} \times 350 = 14 ∴ MD (\overline{x}) = \frac{1}{N} i = 1 \sum 5 f_{i} ∣ x_{i} - \overline{x} ∣ = \frac{1}{25} \times 158 = 6.32$

$x_{i}$	10	30	50	70	90
$f_{i}$	4	24	28	16	8

Sol.

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$∥ x_{i} - \overline{x} ∥$	$f_{i} ∥ x_{i} - \overline{x} ∥$
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
	80	4000		1280

$N = i = 1 \sum 5 f_{i} = 80, i = 1 \sum 5 f_{i} x_{i} = 4000 ∴ \overline{x} = \frac{1}{N} i = 1 \sum 5 f_{i} x_{i} = \frac{1}{80} \times 4000 = 50 MD (\overline{x}) \frac{1}{N} i = 1 \sum 5 f_{i} ∣ x_{i} - \overline{x} ∣ = \frac{1}{80} \times 1280 = 16$

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

$x_{i}$	5	7	9	10	12	15
$f_{i}$	8	6	2	2	2	6

Sol. The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

$x_{i}$	$f_{i}$	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 $1 3^{th}$ and $1 4^{th}$ observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

$∴$ Median $= \frac{1 3 ^{th} observation + 1 4 ^{th} observation}{2}$

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

The absolute values of the deviations from median, i.e. $∣ x_{i} - M ∣$ are

$∥ x_{i} - M ∥$	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

$i = 1 \sum 6 f_{i} = 26 and i = 1 \sum 6 f_{i} ∣ x_{i} - M ∣ = 84 M.D. (M) = \frac{1}{N} i = 1 \sum 6 f_{i} ∣ x_{i} - M ∣ = \frac{1}{26} \times 84 = 3.23$

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

Sol. The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

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

Here, $N = 29$ , which is odd.

∴ Median $= (\frac{29 + 1}{2})^{th}$ observation $= 1 5^{th}$ observation

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

$∴ Median = 30$

The absolute values of the deviations from median, i.e. $∣ x_{i} - M ∣$ are

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

$∴ i = 1 \sum 5 f_{i} = 29, i = 1 \sum 5 f_{i} ∣ x_{i} - M ∣ = 148 M . D . (M) = \frac{1}{N} i = 1 \sum 5 f_{i} ∣ x_{i} - M ∣ = \frac{1}{29} \times 148 = 5.1$

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

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 following table is formed.

Income per day	No. of persons $f_{i}$	Midpoint $x_{i}$	$f_{i} x_{i}$	$∥ x_{i} - \overline{x} ∥$	$f_{i} \times x_{i} - \overline{x}$
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
	50		17900		7896

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

$i = 1 \sum 8 f_{i} x_{i} = 17900 ∴ \overset{x}{ˉ} = \frac{1}{N} i = 1 \sum 8 f_{i} x_{i} = \frac{1}{50} \times 17900 = 358$

M.D.

$(\overset{x}{ˉ}) = \frac{1}{N} i = 1 \sum 8 f_{i} ∣ x_{i} - \overset{x}{ˉ} ∣ = \frac{1}{50} \times 7896 = 157.92$

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

Sol. The following table is formed.

Height in cms	No. of boys $f_{i}$	Midpoint $x_{i}$	$f_{i} x_{i}$	$∥ x_{i} - \overline{x} ∥$	$f_{i} ∥ x_{i} - \overline{x} ∥$
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

$Here, N = i = 1 \sum 6 f_{i} = 100, i = 1 \sum 6 f_{i} x_{i} = 12530 ∴ \overset{x}{ˉ} = \frac{1}{N} i = 1 \sum 6 f_{i} x_{i} = \frac{1}{100} \times 12530 = 125.3 M.D. (\overset{x}{ˉ}) = \frac{1}{N} i = 1 \sum 6 f_{i} ∣ x_{i} - \overset{x}{ˉ} ∣ = \frac{1}{100} \times 1128.8 = 11.28$

Find the mean deviation about the median for the following 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 following table is formed.

Marks	No. of girls $f_{i}$	(c.f.)	Midpoint $x_{i}$	$∥ x_{i} -$ med. $∥$	$f_{i} ∣ x_{i} -$ med.\|
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 the $(\frac{N}{2})^{th}$ or $2 5^{th}$ item is 20-30. Therefore, 20-30 is the median class. It is known that,

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

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

$∴ Median = 20 + \frac{25 - 14}{14} \times 10 = 20 + \frac{110}{14} = 20 + 7.85 = 27.85$

Thus, mean deviation about the median is given by,

$M.D. (M) = \frac{1}{N} i = 1 \sum 6 f_{i} ∣ x_{i} - M ∣ = \frac{1}{50} \times 517.1 = 10.34$

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

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 formed as follows.

Marks	No. of girls $f_{i}$	(c.f.)	Midpoint $x_{i}$	$∣ x_{i} -$ med.\|	$f_{i} ∣ x_{i} -$ med.\|
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
	100				735

The class interval containing the $(\frac{N}{2})^{th}$ or $5 0^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class. It is known that,

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

Here, $l = 3.5, C = 37, f = 26, h = 5$ and $N = 100$

$∴ Median = 35.5 + \frac{50 - 37}{26} \times 5 = 35.5 + \frac{13 \times 5}{26} = 35.5 + 2.5 = 38$

Thus, mean deviation about the median is given by, M.D.

$(M) = \frac{1}{N} i = 1 \sum 8 f_{i} ∣ x_{i} - M ∣ = \frac{1}{100} \times 735 = 7.35$

EXERCISE - 13.2

Find the mean and variance for the data in Q. 1 to 5

6, 7, 10, 12, 13, 4, 8, 12 Sol. 6, 7, 10, 12, 13, 4, 8, 12 Mean, $\overline{x} = \frac{\sum _{i = 1}^{8} x _{i}}{n}$

$= \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = 9$

The following table is obtained.

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

$= \frac{1}{8} \times 74 = 9.25$

n natural numbers. Sol. The mean of first n natural numbers is calculated as follows

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

Variance $(σ^{2}) = \frac{1}{n} \sum_{i = 1}^{n} (x_{i} - \overline{x})^{2}$

$= \frac{1}{n} i = 1 \sum n [x_{i} - (\frac{n + 1}{2})]^{2} = \frac{1}{n} i = 1 \sum n x_{i}^{2} - \frac{1}{n} i = 1 \sum n 2 (\frac{n + 1}{2}) x_{i} + \frac{1}{n} i = 1 \sum n (\frac{n + 1}{2})^{2} = \frac{1}{n} \frac{n ( n + 1 ) ( 2 n + 1 )}{6} - (\frac{n + 1}{n}) [\frac{n ( n + 1 )}{2}] + \frac{( n + 1 ) ^{2}}{4 n} \times n = \frac{( n + 1 ) ( 2 n + 1 )}{6} - \frac{( n + 1 ) ^{2}}{2} + \frac{( n + 1 ) ^{2}}{4} = \frac{( n + 1 ) ( 2 n + 1 )}{6} - \frac{( n + 1 ) ^{2}}{4} = (n + 1) [\frac{4 n + 2 - 3 n - 3}{12}] = \frac{( n + 1 ) ( n - 1 )}{12} = \frac{n ^{2} - 1}{12}$

First 10 multiples of 3

Sol. The first 10 multiples of 3 are

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

Here, number of observations, $n = 10$

$Mean, \overline{x} = \frac{\sum _{i = 1}^{10} x _{i}}{10} = \frac{165}{10} = 16.5$

The following table is obtained

$x_{i}$	$(x_{i} - \overline{x})$	$(x_{i} - \overline{x})^{2}$
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

$Variance (σ^{2}) = \frac{1}{n} i = 1 \sum 10 (x_{i} - \overset{x}{ˉ})^{2} = \frac{1}{10} \times 742.5 = 74.25$

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

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

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

Here, $N = 40, \sum_{i = 1}^{7} f_{i} x_{i} = 760$

$∴ \overset{x}{ˉ} = \frac{\sum _{i = 1}^{7} f _{i} x _{i}}{N} Variance (σ^{2}) = \frac{760}{40} = 19 = \frac{1}{n} i = 1 \sum 7 f_{i} (x_{i} - \overset{x}{ˉ})^{2} = \frac{1}{40} \times 1736 = 43.4$

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

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

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$x_{i} - \overline{x}$	$(x_{i} - \overline{x})^{2}$	$f_{i} (x_{i} - \overline{x})^{2}$
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
	22	2200			640

Here, $N = 22, \sum_{i = 1}^{7} f_{i} x_{i} = 2200$

$∴ \overline{x} = \frac{\sum _{i = 1}^{7} f _{i} x _{i}}{N} = \frac{1}{22} \times 2200 = 100$

Variance $(σ^{2}) = \frac{1}{n} \sum_{i = 1}^{7} f_{i} (x_{i} - \overline{x})^{2}$

$= \frac{1}{22} \times 640 = 29.09$

Find 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.

$x_{i}$	$f_{i}$	$y_{i} = \frac{x _{i} - 64}{1} x_{i}$	$y_{i}^{2}$	$f_{i} y_{i}$	$f_{i} y_{i}^{2}$
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			0	286

Let Assumed mean be $A = 64$ Mean,

$\overset{x}{ˉ} = A + \frac{\sum _{i = 1}^{9} f _{i} y _{i}}{N} \times h = 64 + \frac{0}{100} \times 1 = 64 + 0 = 64$

Variance

$(σ^{2}) = \frac{h ^{2}}{N ^{2}} N i = 1 \sum 9 f_{i} y_{i}^{2} - (i = 1 \sum 9 f_{i} y_{i})^{2} = \frac{1}{10 0 ^{2}} [100 \times 286 - 0] = 2.86$

∴ Standard deviation $(σ)$

$= 2.86 = 1.69$

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

Classes	030	3060	6090	90120	120150	150180	180210
$f_{i}$	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		220		2	76

Let Assumed mean be $A = 105$ Mean,

$\overline{x} = A + \frac{\sum _{i = 1}^{7} f _{i} y _{i}}{N} \times h = 105 + \frac{2}{30} \times 30 = 105 + 2 = 107$

Variance

$(σ^{2}) = \frac{h ^{2}}{N ^{2}} N i = 1 \sum 7 f_{i} y_{i}^{2} - (i = 1 \sum 7 f_{i} y_{i})^{2} = \frac{( 30 ) ^{2}}{( 30 ) ^{2}} [30 \times 76 - (2)^{2}] = 2280 - 4 = 2276$

Classes	0-10	10-20	20-30	30-40	40-50
$f_{i}$	5	8	15	16	6

Sol.

Class	$f_{i}$	$x_{i}$	$y_{i} = \frac{x _{i} - 25}{10}$	$y_{i}^{2}$	$f_{i} y_{i}$	$f_{i} y_{i}^{2}$
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

Let Assumed mean be $A = 25$ Mean, $\overline{x} = A + \frac{\sum _{i = 1}^{5} f _{i} y _{i}}{N} \times h$

$= 25 + \frac{10}{50} \times 10 = 25 + 2 = 27$

Variance $(σ^{2}) = \frac{h ^{2}}{N ^{2}} [N \sum_{i = 1}^{5} f_{i} y_{i}^{2} - (\sum_{i = 1}^{5} f_{i} y_{i})^{2}]$

$= \frac{( 10 ) ^{2}}{( 50 ) ^{2}} [50 \times 68 - (10)^{2}] = \frac{1}{25} [3400 - 100] = \frac{3300}{25} = 132$

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

Height in cms	No. 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 Interval	$f_{i}$	$x_{i}$	$y_{i} = \frac{x _{i} - 92.5}{5}$	$y_{i}^{2}$	$f_{i} y_{i}$	$f_{i} y_{i}^{2}$
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

Let Assumed mean be $A = 92.5$ Mean, $\overline{x} = A + \frac{\sum _{i = 1}^{9} f _{i} y _{i}}{N} \times h$

$= 92.5 + \frac{6}{60} \times 5 = 92.5 + 0.5 = 93$

∴ Standard deviation $(σ) = 105.58 = 10.27$

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

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

Sol.

Class Interval	$f_{i}$	$x_{i}$	$y_{i} = \frac{x _{i} - 42.5}{4}$	$y_{i}^{2}$	$f_{i} y_{i}$	$f_{i} y_{i}^{2}$
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

Here, $N = 100, h = 4$ Let the assumed mean, A be 42.5

$Mean, \overset{x}{ˉ} = A + \frac{\sum _{i = 1}^{5} f _{i} y _{i}}{N} \times h = 42.5 + \frac{25}{100} \times 4 = 43.5$

Variance $(σ^{2}) = \frac{h ^{2}}{N ^{2}} [N \sum_{i = 1}^{5} f_{i} y_{i}^{2} - (\sum_{i = 1}^{5} f_{i} y_{i})^{2}]$

$= \frac{16}{10000} [100 \times 199 - (25)^{2}] = \frac{16}{10000} [19900 - 625] = \frac{16}{10000} \times 19275 = 30.84$

∴ Standard deviation $(σ) = 5.55$

MISCELLANEOUS EXERCISE

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. Sol. Let the remaining two observations be x and y. Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y. Mean, $\overline{x} = \frac{6 + 7 + 10 + 12 + 12 + 13 + x + y}{8} = 9$

$\Rightarrow \Rightarrow 60 + x + y = 72 x + y = 12$

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

$9.25 = \frac{1}{8} [(- 3)^{2} + (- 2)^{2} + (1)^{2} + (3)^{2} + (3)^{2} + (4)^{2} + x^{2} + y^{2} - 2 \times 9 (x + y) + 2 \times (9)^{2}]$

From (1), we obtain

$x^{2} + y^{2} + 2 x y = 144$

From (2) and (3), we obtain

$2 x y = 64$

Subtracting (4) from (2), we obtain

$\Rightarrow x^{2} + y^{2} - 2 x y = 80 - 64 = 16 x - y = \pm 4$

Therefore, from (1) and (5), we obtain $x = 8$ and $y = 4$ , when $x - y = 4$ $x = 4$ and $y = 8$ , when $x - y = - 4$ Thus, the remaining observations are 4 and 8.

2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations. Sol. Let the remaining two observations be x and y. The observations are 2, 4, 10, 12, 14, x, y. Mean, $\overline{x} = \frac{2 + 4 + 10 + 12 + 14 + x + y}{7} = 8$

$\Rightarrow \Rightarrow 56 = 42 + x + y x + y = 14$

Variance $= 16 = \frac{1}{n} \sum_{i = 1}^{7} (x_{i} - \overline{x})^{2}$

$16 = \frac{1}{7} [(- 6)^{2} + + (- 4)^{2} + 2^{2} + 4^{2} + 6^{2} x^{2} + y^{2} - 2 \times 8 (x + y) + 2 \times 8^{2}]$

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

$16 = \frac{1}{7} [108 + x^{2} + y^{2} - 224 + 128]$ $16 = \frac{1}{7} [12 + x^{2} + y^{2}]$

$\Rightarrow x^{2} + y^{2} = 112 - 12 = 100$

From (1), we obtain

$x^{2} + y^{2} + 2 x y = 196$

From (2) and (3), we obtain

$\Rightarrow 2 x y 2 x y = 196 - 100 = 96$

Subtracting (4) from (2), we obtain

$\Rightarrow \Rightarrow x^{2} + y^{2} - 2 x y = 100 - 96 (x - y)^{2} = 4 x - y = \pm 2$

Therefore, from (1) and (5), we obtain $x = 8$ and $y = 6$ , when $x - y = 2$ $x = 6$ and $y = 8$ , when $x - y = - 2$ Thus, the remaining observations are 6 and 8.

The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations. Sol. Let the observations be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ and $x_{6}$ . It is given that mean is 8 and standard deviation is 4.

$Mean, \overset{x}{ˉ} = \frac{x _{1} + x _{2} + x _{3} + x _{4} + x _{5} + x _{6}}{6} = 8$

If each observation is multiplied by 3 and the resulting observations are $y_{i}$ , then

$y_{i} = 3 x_{i}, i.e. x_{i} = \frac{1}{3} y_{i} for i = 1 to 6$

$∴$ New mean, $\overline{y} = \frac{y _{1} + y _{2} + y _{3} + y _{4} + y _{5} + y _{6}}{6}$

$= \frac{3 ( x _{1} + x _{2} + x _{3} + x _{4} + x _{5} + x _{6} )}{6} = 3 \times 8 = 24 \dots\dots [Using (1)]$

Standard deviation, $σ = \frac{1}{n} \sum_{i = 1}^{6} (x_{i} - \overline{x})^{2}$

$∴ ∴ (4)^{2} = \frac{1}{6} \sum_{i = 1}^{6} (x_{i} - \overline{x})^{2} \sum_{i = 1}^{6} (x_{i} - \overline{x})^{2} = 96$

From (1) and (2), it can be observed that,

$\overline{y} = 3 \overline{x} \Rightarrow \overline{x} = \frac{1}{3} \overline{y}$

Substituting the values of $x_{i}$ and $\overline{x}$ in (2), we obtain

$\sum_{i = 1}^{6} (\frac{1}{3} y_{i} - \frac{1}{3} \overline{y})^{2} = 96 \Rightarrow \sum_{i = 1}^{6} (y_{i} - \overline{y})^{2} = 864$

Therefore, variance of new observations

$= (\frac{1}{6} \times 864) = 144$

Hence, the standard deviation of new observations is $144 = 12$

Given that $\overline{x}$ is the mean and $σ^{2}$ is the variance of n observations $x_{1}, x_{2} \dots\dots x_{n}$ . Prove that the mean and variance of the observations $ax_{1}, ax_{2} \dots .. ax_{n}$ are $a \overline{x}$ and $a^{2} σ^{2}$ , respectively ( $a \neq = 0$ ). Sol. The given n observations are $x_{1}, x_{2} \dots\dots x_{n}$ .

$Mean = \overset{x}{ˉ} Variance = σ^{2} ∴ σ^{2} = \frac{1}{n} i = 1 \sum n (x_{i} - \overset{x}{ˉ})^{2}$

If each observation is multiplied by a and the new observations are $y_{i}$ , then

$y_{i} = a x_{i} i.e. x_{i} = \frac{1}{a} y_{i} ∴ \overset{y}{ˉ} = \frac{1}{n} \sum_{i = 1}^{n} y_{i} = \frac{1}{n} \sum_{i = 1}^{n} a x_{i} = \frac{a}{n} \sum_{i = 1}^{n} x_{i} = a \overset{x}{ˉ} (\overset{x}{ˉ} = \frac{1}{n} \sum_{i = 1}^{n} x_{i})$

Therefore, mean of the observations, $ax_{1}, ax_{2} \dots ax_{n}$ , is $a \overline{x}$ . Substituting the values of $x_{i}$ and $\overline{x}$ in (1), we obtain

$σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} (\frac{1}{a} y_{i} - \frac{1}{a} \overset{y}{ˉ})^{2}$

$\Rightarrow a^{2} σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} (y_{i} - \overline{y})^{2}$ Thus, the variance of the observations, $ax_{1}, ax_{2} \dots .. ax_{n}$ , is $a^{2} σ^{2}$ .

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

Sol.

(i) Number of observations $(n) = 20$ Incorrect mean = 10 Incorrect standard deviation $= 2$

$\overline{x} = \frac{1}{n} \sum_{i = 1}^{20} x_{i}$

$\Rightarrow 10 = \frac{1}{20} \sum_{i = 1}^{20} x_{i} \Rightarrow \sum_{i = 1}^{20} x_{i} = 200$

That is, incorrect sum of observations $= 200$ Correct sum of observations $= 200 - 8 = 192$

$Correct mean = \frac{Correct sum}{19} = \frac{192}{19} = 10.1$

Standard deviation $(σ)$

$= \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - \frac{1}{n ^{2}} (\sum_{i = 1}^{n} x_{i})^{2} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - (\overset{x}{ˉ})^{2}$

Standard deviation $(σ)$

$= \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - \frac{1}{n ^{2}} (\sum_{i = 1}^{n} x_{i})^{2} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - (\overset{x}{ˉ})^{2}$

$\Rightarrow 2 = \frac{1}{20} Incorrect \sum_{i = 1}^{n} x_{i}^{2} - (10)^{2}$ $\Rightarrow 4 = \frac{1}{20} Incorrect \sum_{i = 1}^{n} x_{i}^{2} - 100$ $\Rightarrow$ Incorrect $\sum_{i = 1}^{n} x_{i}^{2} = 2080$ ∴ Correct $\sum_{i = 1}^{n} x_{i}^{2}$

$= Incorrect i = 1 \sum n x_{i}^{2} - (8)^{2} + (12)^{2} = 2080 - 64 + 144 = 2160$

$\Rightarrow 2 = \frac{1}{20} Incorrect \sum_{i = 1}^{n} x_{i}^{2} - (10)^{2}$

$\Rightarrow 4 = \frac{1}{20} Incorrect \sum_{i = 1}^{n} x_{i}^{2} - 100$

$\Rightarrow Incorrect \sum_{i = 1}^{n} x_{i}^{2} = 2080$

$∴ Correct i = 1 \sum n x_{i}^{2} = Incorrect i = 1 \sum n x_{i}^{2} - (8)^{2} = 2080 - 64 = 2016$

Correct standard deviation

$= \frac{Correct \sum x _{i}^{2}}{n} - (Correct mean)^{2} = \frac{2160}{20} - (10.2)^{2} = 108 - 104.04 = 3.96 = 1.98$

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

Sol. Number of observations $(n) = 100$ Incorrect mean $(\overline{x}) = 20$ Incorrect standard deviation $(σ) = 3$

$\Rightarrow 20 = \frac{1}{100} i = 1 \sum 100 x_{i} \Rightarrow i = 1 \sum 100 x_{i} = 20 \times 100 = 2000$

(ii) When 8 is replaced by 12, Incorrect sum of observations = 200 ∴ Incorrect sum of observations = 2000 ∴ Correct sum of observations

$= 200 - 8 + 12 = 204$

⇒ Correct sum of observations

$= 2000 - 21 - 21 - 18 = 2000 - 60 = 1940$

∴ Correct mean $= \frac{Correct sum}{20} = \frac{204}{20} = 10.2$ $∴$ Correct mean $= \frac{Correct sum}{100 - 3} = \frac{1940}{97} = 20$

Standard deviation $(σ)$

$= \frac{1}{n} i = 1 \sum n x_{i}^{2} - \frac{1}{n ^{2}} (i = 1 \sum n x_{i})^{2} = \frac{1}{n} i = 1 \sum n x_{i}^{2} - (\overset{x}{ˉ})^{2} \Rightarrow 3 = \frac{1}{100} Incorrect i = 1 \sum n x_{i}^{2} - (20)^{2} \Rightarrow Incorrect x_{i}^{2} \sum = 100 (9 + 400) = 40900 \Rightarrow Correct i = 1 \sum n x_{i}^{2} = Incorrect i = 1 \sum n x_{i}^{2} - (21)^{2} - (21)^{2} - (18)^{2} = 40900 - 441 - 441 - 324 = 39694$

∴ Correct standard deviation

$= \frac{Correct \sum x _{i}^{2}}{n} - (Correct mean)^{2} = \frac{39694}{97} - (20)^{2} = 409.216 - 400 = 9.216 = 3.036$

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Access chapter-wise NCERT Class 11 Maths solutions with correct answers, important formulas, and easy explanations. Understand concepts better, solve questions with confidence, and improve your problem-solving skills for exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 14	Probability

4.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 13

Exercise	Number of Questions	Important Concepts Covered
Exercise 13.1	12 Questions & Solutions	Mean, median, mode, measures of central tendency, grouped and ungrouped data
Exercise 13.2	10 Questions & Solutions	Range, quartile deviation, mean deviation, standard deviation, grouped and ungrouped data
Miscellaneous Exercise	6 Questions & Solutions	Mixed questions on central tendency, dispersion, grouped and ungrouped data, frequency distribution tables

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 13

Organized Tabular Solutions: The solutions present data in clear, multi-column tables, showing exactly how to calculate f_ix_i, (x_i - \bar{x}), and f_i(x_i - \bar{x})^2, which minimizes confusion.
Mean Deviation from Mean and Median: Learn how to calculate mean deviation about the mean and median with clear formulas and step-by-step calculations.
Variance and Standard Deviation: Understand how to calculate variance and standard deviation for different sets of data, including grouped frequency distributions.
Step-Deviation Method: Learn how to use the step-deviation method to simplify calculations when working with large or complex data sets.
Coefficient of Variation: Understand how to calculate the coefficient of variation and use it to compare the consistency of two data sets, such as the performance of two players.
Accurate Statistical Calculations: The solutions explain each calculation carefully, including squaring deviations and adding values, to help students avoid common calculation mistakes.
Prepared by ALLEN Subject Experts: These NCERT Solutions are prepared by ALLEN subject experts with accurate mathematical methods and alignment with the latest NCERT syllabus.
Complete NCERT Exercise Coverage: All questions from the NCERT exercises, including the miscellaneous exercise, are covered with clear, step-by-step solutions to help students understand and apply the concepts of Statistics.

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NCERT Solutions Class 11 Maths Chapter 13 – Statistics

1.0Class 11 Maths Chapter 13: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 13

EXERCISE - 13.1

EXERCISE - 13.2

MISCELLANEOUS EXERCISE

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

4.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 13

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 13

On this page

$x_{i}$	$f_{i}$	$y_{i} = \frac{x _{i} - 64}{1} x_{i}$	$y_{i}^{2}$	$f_{i} y_{i}$	$f_{i} y_{i}^{2}$
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			0	286

$x_{i}$	$f_{i}$	$y_{i} = \frac{x _{i} - 64}{1} x_{i}$	$y_{i}^{2}$	$f_{i} y_{i}$	$f_{i} y_{i}^{2}$
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			0	286

$x_{i}$	$f_{i}$	$y_{i} = \frac{x _{i} - 64}{1} x_{i}$	$y_{i}^{2}$	$f_{i} y_{i}$	$f_{i} y_{i}^{2}$
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			0	286