NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Exercise 15.2

June 11, 2022

NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Exercise 15.2

Textbook	NCERT
class	Class – 11th
Subject	Mathematics
Chapter	Chapter – 15
Chapter Name	Statistics
grade	Class 11th Maths solution
Medium	English
Source	last doubt

NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Exercise 15.2

?Chapter – 15?

✍Statistics✍

?Exercise 15.2?

1. Find the mean and variance for each of the data in Exercise 1 to 5.
1. 6, 7, 10, 12, 13, 4, 8, 12

?‍♂️Solution:

So, x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
Let us make the table of the given data and append other columns after calculations.

X_i	Deviations from mean(x_i – x̅)	(x_i – x̅)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

We know that Variance,

∑ (x_i– x̅ )² / n

= 74/8

= 9.25

2. First n natural numbers

?‍♂️Solution: First n natural numbers: 1, 2, 3, ….., n

Mean x̅ = 1+2+3+….+ 1/n × n(n+1)/2

= n+1/2 ….{sum of first natural numbers n(n+1)/2}

∑ x ²_i = 1 ² +2 ² +3 ² +….+n ²

= n ( n + 1) (2n + 1) / 6

Variance = ∑ (x_i– x̅ )² / n = 1/n²[ n ∑x²_i– (∑x_i)²

= 1/n[n × n (n + 1) (2n + 1)/6 – n²(n+1)²/4]

= 1/12[2(n+1)(2n+1) – 3(n+1)²]

= n+1/12 [2(2n+1)-3(n+1)]

= n+1/12[4n+2-3n-3]

= (n+1)(n-1)/12

= n²– 1/12

3. First 10 multiples of 3

?‍♂️Solution:

So, x̅ = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10
= 165/10
= 16.5
Let us make the table of the data and append other columns after calculations

X_i	Deviations from mean(x_i – x̅)	(x_i – x̅)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

Then, Variance

σ²= h²/h²[n ∑y²_i– (∑y_i)²]

= 9/100[10×85-25]

= 9/100[850-25]

= 9×825/100

= 7425/100

= 74.25

अतः माध्य = 16.5, प्रसरण = 74.25

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

?‍♂️Solution: Let us make the table of the given data and append other columns after calculations.

X_i	f_i	f_ix_i	Deviations from mean(x_i – x̅)	(x_i – x̅)²	f_i(x_i – x̅)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
	N = 40	760			1736

Mean = x̅ = 760/40 = 19

Variance = σ²= ∑f_i(x_i – x̅)²/N

= 1736/40

= 43.3

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

?‍♂️Solution: Let us make the table of the given data and append other columns after calculations.

X_i	f_i	f_ix_i	Deviations from mean(x_i – x̅)	(x_i – x̅)²	f_i(x_i – x̅)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 = x̅ = A + ∑y_i/n

= 98 + 44/22

= 98 + 2

= 100

Variance σ²= 1/N²[∑f_iy_i – (∑f_iy_i)²]

= 1/(22)²[22×728 – 44 × 44]

= 1/22[728-88]

= 640/22

= 320/11

= 29.09

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: Let the assumed mean A = 64. Here h = 1

We obtain the following table from the given data.

Let the assumed mean A = 64

and y_i = x_i – 64

X_i	Frequency f_i	Y_i = (x_i – A)/h	Y_i²	f_iy_i	f_iy_i²
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
				0	286

Mean, x̅ = A + ∑f_iy_i/N

= 64 + 0

= 64

Then, variance,

σ²= 1/N²[∑f_iy_i – (∑f_iy_i)²]

= 1/(100)²[100×286 – 0]

= 286/100

= √2.86

Therefore, Standard variance = 2.886
= 1.691
Mean = 64 and Standard variance = 1.691

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

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

?‍♂️Solution: Let us make the table of the given data and append other columns after calculations.

the assumed mean A = 105, class interval h = 30

y_i = x_i – A / h = x_i – 10 /30

Classes	Frequency f_i	Mid – pointsx_i	f_ix_i	(x_i – x̅)	(x_i – x̅)²	f_i(x_i – x̅)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 x̅ = A + (∑f_iy_i/N) x h

= 105+2/30×30

= 107

Variance σ²= h²/h²[n ∑f_iy²_i– (∑f_iy_i)²]

= 30×30/30×30[30×76-2²]

= 2280 − 4

= 2276

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

?‍♂️Solution: Let us make the table of the given data and append other columns after calculations.

y_i = x_i – A / h = x_i – 25 /10

Classes	Frequency f_i	Mid – pointsx_i	f_ix_i	(x_i – x̅)	(x_i – x̅)²	f_i(x_i – x̅)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 x̅ = A + (∑f_iy_i/N) x h

= 25+(10/50)×10

= 25 + 2

= 27

Variance σ²= h²/h²[n ∑f_iy²_i– (∑f_iy_i)²]

= 100/2500[50×68-100]

= 50/25[68-2]

= 2 × 66

= 132

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
Frequencies	3	4	7	7	15	9	6	6	3

?‍♂️Solution: Let the assumed mean, A = 92.5 and h = 5

Let us make the table of the given data and append other columns after calculations.

Height (class)	Number of children Frequency f_i	MidpointX_i	Y_i = (x_i – A)/h	Y_i²	f_iy_i	f_iy_i²
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
	N = 60				6	254

Mean, x̅ = A + (∑f_iy_i/N) x h

= 9.25+6/60×5

= 92.5 + 0.5

= 93

Variance,

σ²= h²/h²[n ∑f_iy²_i– (∑f_iy_i)²]

= 25/3600[60×254-36]

= 12/144[5×254-3]

= 112[1270-3]

= 1267/12

= 105.58

Standard Deviation, =

σ = √105.58

= 10.28

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: Let the assumed mean, A = 42.5 and h = 4

Let us make the table of the given data and append other columns after calculations.

y_i = x_i – 42.5 /4

Height (class)	Number of children(Frequency f_i)	MidpointX_i	Y_i = (x_i – A)/h	Y_i²	f_iy_i	f_iy_i²
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

Mean, x̅ = A + (∑f_iy_i/N) x h

= 42.5+25100×4

= 42.5 + 1

= 43.5

Variance σ²= h²/h²[n ∑f_iy²_i– (∑f_iy_i)²]

= 16/(100)²[100×199-(25)²]

= 16×25/100×100[4×199-25]

= 1/25[796-25]

= 771/25

= 30.84

Standard Deviation σ = √30.48 = 5.56