Class 11 NCERT Solutions- Chapter 15 Statistics - Exercise 15.2

Chapter 15 of Class 11 NCERT focuses on Statistics a crucial branch of mathematics that deals with the data collection, analysis, and interpretation. Exercise 15.2 dives deeper into the various statistical methods including the computation of measures of the dispersion such as range, variance, and standard deviation for both the grouped and ungrouped data. These concepts form the backbone of the data analysis and are essential for students to understand real-world data behavior.

Statistics

Statistics is a mathematical science that involves methods to collect, classify, analyze, and interpret data. It helps in making decisions based on the data, predicting future trends and solving complex problems in various fields like economics, biology, and engineering. In this chapter, students learn different tools for summarizing and describing data efficiently. The measures of central tendency and dispersion are explored giving a comprehensive view of the data distribution.

Find the mean and variance for each of the data in Exercise 1 to 5.

Question 1. 6, 7, 10, 12, 13, 4, 8, 12

Solution:

We know,
Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}
So, \bar{x} = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
x_i
Deviations from mean
(x_i- x')
(xi - x')²
6 6 – 9 = -3 9
7 7 – 9 = -2 4
10 10 – 9 = 1 1
12 12 – 9 = 3 9
13 13 – 9 = 4 16
4 4 – 9 = – 5 25
8 8 – 9 = – 1 1
12 12 – 9 = 3 9
74
\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2
σ²= (1/8) × 74
= 9.2
Therefore, Mean = 9 and Variance = 9.25

x_i	Deviations from mean (x_i- x')	(xi - x')²
6	6 – 9 = -3	9
7	7 – 9 = -2	4
10	10 – 9 = 1	1
12	12 – 9 = 3	9
13	13 – 9 = 4	16
4	4 – 9 = – 5	25
8	8 – 9 = – 1	1
12	12 – 9 = 3	9
		74

Question 2. First n natural numbers

Solution:

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}
\bar{x} = ((n(n + 1))2)/n
= (n + 1)/2
\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2
On substituting the value of mean,
= \frac{1}{n}\sum_{i=1}^{n}(x_i - \frac{n+1}{2})^2 \\ = \frac{1}{n}\sum_{i=1}^{n}(x_i)^2 - \frac{1}{n}\sum_{i=1}^{n}2x_i(\frac{n+1}{2})+\frac{1}{n}\sum_{i=1}^{n}(\frac{n+1}{2})^2 Substituting values of Summation \\ = \frac{1}{n}\frac{n(n+1)(2n+1)}{6}-\frac{n+1}{n}[\frac{n(n+1)}{2}]+\frac{(n+1)^2}{4n}×n
On extracting common values, we have,
= (n+1)[\frac{4n+2-3n-3}{12}] \\ = \frac{(n+1)(n-1)}{12}
σ² = (n² – 1)/12
Mean = (n + 1)/2 and Variance = (n² – 1)/12

Question 3. First 10 multiples of 3

Solution:

The required multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
We know,
Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}
So, \bar{x} = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10
= 165/10
= 16.5
x_i
Deviations from mean
(x_i- x')
(x_i - x')²
3 3 – 16.5 = -13.5 182.25
6 6 – 16.5 = -10.5 110.25
9 9 – 16.5 = -7.5 56.25
12 12 – 16.5 = -4.5 20.25
15 15 – 16.5 = -1.5 2.25
18 18 – 16.5 = 1.5 2.25
21 21 – 16.5 = – 4.5 20.25
24 24 – 16.5 = 7.5 56.25
27 27 – 16.5 = 10.5 110.25
30 30 – 16.5 = 13.5 182.25
742.5
\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2
= (1/10) × 742.5
= 74.25
Therefore, Mean = 16.5 and Variance = 74.25

x_i	Deviations from mean (x_i- x')	(x_i - x')²
3	3 – 16.5 = -13.5	182.25
6	6 – 16.5 = -10.5	110.25
9	9 – 16.5 = -7.5	56.25
12	12 – 16.5 = -4.5	20.25
15	15 – 16.5 = -1.5	2.25
18	18 – 16.5 = 1.5	2.25
21	21 – 16.5 = – 4.5	20.25
24	24 – 16.5 = 7.5	56.25
27	27 – 16.5 = 10.5	110.25
30	30 – 16.5 = 13.5	182.25
		742.5

Question 4.

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

Solution:

x_i	f_i	f_ix_i	x_i - x'	(x_i - x')²	f_i(x_i- x')²
6	2	12	6 – 19 = 13	169	338
10	4	40	10-19 = -9	81	324
14	7	98	14-19 = -5	25	175
18	12	216	18-19 = -1	1	12
24	8	192	24-19 = 5	25	200
28	4	112	28-19 = 9	81	324
30	3	90	30-19 = 11	121	363
					1736

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}
\bar{x} = 760/40
= 19
Also,
\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2
= (1/40) × 1736
= 43.4

Question 5.

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

Solution:

x_i	f_i	f_ix_i	x_i- x'	(x_i - x')²	f_i(x_i - x')²
92	3	276	92-100 = -8	64	192
93	2	186	93-100 = -7	49	98
97	3	291	97-100 = -3	9	27
98	2	196	98-100 = -2	4	8
102	6	612	102-100 = 2	4	24
104	3	312	104-100 =4	16	48
109	3	327	109-100 = 9	81	243
	N = 22	2200			640

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}
\bar{x} = 2200/22
= 100
\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2
= (1/22) × 640
= 29.09
Therefore, Mean = 100 and Variance = 29.09

Question 6. 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

Solution:

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h
Where A = 64, h = 1
So, \bar{x} = 64 + ((0/100) × 1)
= 64 + 0
= 64
Then, variance,
\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]
σ² = (1²/100²) [100(286) – 0²]
= (1/10000) [28600 – 0]
= 28600/10000
= 2.86
Hence, standard deviation = σ = √2.886
= 1.691
Therefore,
Mean = 64 and Standard Deviation = 1.691

Question 7.

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

Solution:

Classes	f_i	x_i	f_ix_i	(x_i- x')	(xi - x')²	f_i(xi - 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
	N = 30		3210			68280

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}
\bar{x} = 3210/30
= 107
\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2
= (1/30) × 68280
= 2276
Therefore, Mean = 107 and Variance = 2276

Question 8.

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

Solution:

Classes	f_i	x_i	f_ix_i	(x_i-x')	(x_i-x')²	f_i(x_i-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
	N = 50		1350			6600

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}
\bar{x} = 1350/50
= 27
\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2
= (1/50) × 6600
= 132
Therefore, Mean = 27 and Variance = 132

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

Heights in cms	70-75	75-80	80-85	85-90	90-95	95-100	100-105	105-110	110-115
Frequencies	3	4	7	7	15	9	6	6	3

Solution:

Height	f_i	X_i	Y_{i =}(X_i-A)/h	Y_i²	f_iy_i	f_iy_i²
70-75	2	72.5	-4	19	-12	48
75-80	1	77.5	-3	9	-12	36
80-85	12	82.5	-2	4	-14	28
85-90	29	87.5	-1	1	-7	7
90-95	25	92.5	0	0	0	0
95-100	12	97.5	1	1	9	9
100-105	10	102.5	2	4	12	24
105-110	4	107.5	3	9	18	54
110-115	5	112.5	4	16	12	48
115-120	N = 60				6	254

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h
Where, A = 92.5, h = 5
So, \bar{x}= 92.5 + ((6/60) × 5)
= 92.5 + 0.5
= 92.5 + 0.5
= 93
Then, Variance,
\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]
Standard deviation = σ = √105.583
= 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, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]

Solution:

Height	f_i	x_i	Y_i = (X_i-A)/h	Y_i²	f_iy_i	fiyi²
32.5-36.5	15	34.5	-2	4	-30	60
36.5-40.5	17	38.5	-1	1	-17	17
40.5-44.5	21	42.5	0	0	0	0
44.5-48.5	22	46.5	1	1	22	22
48.5-52.5	25	50.5	2	4	50	100
	N=100				25	199

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h
Where, A = 42.5, h = 4
\bar{x} = 42.5 + (25/100) × 4
= 42.5 + 1
= 43.5
Then, Variance,
\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]
σ² = (4²/100²)[100(199) – 25²]
On solving, we get,
= (1/625) [19900 – 625]
= 19275/625
= 771/25
= 30.84
Hence, standard deviation = σ = √30.84
= 5.553

Conclusion

Chapter 15 helps students understand how to calculate and interpret the measures of the dispersion in statistics. Exercise 15.2 focuses on practical problems related to the variance and standard deviation reinforcing the importance of data spread and variability in the real-life scenarios. Mastering these concepts is essential for analyzing data effectively which is useful in the various academic and professional disciplines.

Class 11 NCERT Solutions- Chapter 15 Statistics - Exercise 15.2

Statistics

Find the mean and variance for each of the data in Exercise 1 to 5.

Question 1. 6, 7, 10, 12, 13, 4, 8, 12

Question 2. First n natural numbers

Question 3. First 10 multiples of 3

Question 4.

Question 5.

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

Question 7.

Question 8.

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

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

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, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]

Conclusion

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