OP Malhotra Class 11 Maths Solutions Chapter 21 Measures of Dispersion Ex 21(b)

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S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Ex 21(b)

Question 1.
(i) Five students secured marks as ; 8, 10, 15, 30, 22. Find the standard deviation.
(ii) For a set of ungrouped values the following sums are found :
n = 15, Σx = 480, Σx² = 15735.
Find the standard deviation.
(iii) The standard deviation of the numbers 2, 3, 11, x is 3 1/2. Calculate the values of x.
Solution:
(i)

x	X2
8	64
10	100
15	225
30	900
22	484
Σx = 85	Σx2 = 1773

(iii)

xi	\(x_i^2\)
2	4
3	9
11	121
x	X2
Σxi = 16 + x	\(\Sigma x_i^2\) = 134 + x2

Question 2.
Calculate the standard deviation and variance for the integers 11,12,13,…, 20.
Solution:
Given observations are 11, 12, 13, …., 20

Question 3.
Find the standard deviation of the following set of numbers :
25, 50, 45, 30, 70, 42, 36, 48, 34, 60
Solution:

xi	\(x_i^2\)
25	625
50	2500
45	2025
30	900
70	4900
42	1764
36	1296
48	2304
34	1156
60	3600
Σxi = 440	\(\Sigma x_i^2\) = 21070

Question 4.
Calculate the possible values of x, if the standard deviation of the numbers 2, 3, 2x and 11 is 3.5.
Solution:

xi	\(x_i^2\)
2	4
3	9
2x	4x2
11	121
Σxi = 16 + 2x	\(x_i^2\) = 134 + 4x2

Question 5.
Calculate the standard deviation for the following distribution :

Class interval	0-4	4-8	8-12	12-16
Frequency	4	8	2	1

Solution:
The table of values is given as under:

Class Interval	Frequency fi	Mid-Marks xi	\(x_i^2\)	fixi	\(f_i x_i^2\)
0-4	4	2	4	8	16
4-8	8	6	36	48	288
8-12	2	10	100	20	200
12-16	1	14	196	14	196
	Σfi = 15	Σxi = 22		Σfixi = 90	Σ\(f_i x_i^2\) = 700

Then by direct method, we have

Question 6.
Calculate the standard deviation of the following data :

Size	4	5	6	7	8	9	10
Frequency	6	12	15	28	29	14	15

Solution:

xi	fi	di = xi – A A = 7	fidi	di2	fi di2
4	6	-3	-18	9	54
5	12	-2	-24	4	48
6	15	-1	-15	1	15
7	28	0	0	0	0
8	29	1	29	1	29
9	14	2	28	4	56
10	15	3	45	9	135
	Σfi = 119 = N		Σ fidi  = 45		Σ fi di2 = 337

Then by short cut method, we have

Question 7.
Calculate the standard deviation of the following data:

Class interval	0-6	6-12	12-18	18-24	24-30	30-36	36-40
Frequency	19	25	36	72	51	43	28

Solution:

Class Intervals	frequency fi	xi	di = xi – A A = 21	fidi	di2	fi di2
0-6	19	3	– 18	-342	324	6156
6-12	25	9	– 12	-300	144	3600
12-18	36	15	-6	-216	36	1296
18-24	72	21	0	0	0	0
24-30	51	27	6	360	36	1836
30-36	43	33	12	516	144	6192
36-40	28	38	17	476	289	8092
	Σfi = 274			Σ fidi  = 440		Σ fi di2 = 27172

By short cut method, we have

Question 8.
Calculate the standard deviation for the following data giving the age distribution of persons.

Age in years	20-30	30-40	40-50	50-60	60-70	70-80	80-90
No. of persons	3	61	132	153	140	51	2

Solution:
Same age be continuous variable

age in (years)	Actual limits	No. of persons fi	xi	di = xi = 54.5	ui = \(\frac{d_i}{2}\) i = 10	fiui	fiui2
20-30	19.5-29.5	3	24.5	-30	-3	-9	27
30-40	29.5 – 39.5	31	34.5	-20	– 1	– 122	244
40-50	39.5 -49.5	132	44.5	– 10	-2	– 132	132
50-60	49.5 – 59.5	153	54.5	0	0	0	0
60-70	59.5 – 69.5	140	64.5	10	1	140	140
70-80	69.5 – 79.5	51	74.5	20	2	102	204
80-90	79.5-89.5	2	84.5	30	3	6	18
		Σfi = 54.2				Σ fiui = – 15	Σ fiui2 = 765

Then by step deviation method, we have

Question 9.
The heights, to the nearest cm, of 30 men are given below :

159	170	174	173	175	160	161	164	163	165
164	171	162	170	177	185	181	180	175	165
186	174	168	168	176	176	165	175	167	180

Using class intervals 155-160, 160-165,… draw up a grouped frequency distribution and use this to estimate the Arithmetic mean and standard deviation.
Solution:
Table of values for given data is given as under :

Class intervals	Class Marks xi	fi	di = xi – A A = 172.5	ui = di/C C = 5	fiui	fiui2
155-160	157.5	1	– 15	-3	-3	9
160-165	162.5	6	– 10	-2	– 12	24
165-170	167.5	6	-5	-1	-6	6
170-175	172.5	6	0	0	0	0
175-180	177.5	6	5	1	6	6
180-185	182.5	3	10	2	6	12
185-190	187.5	2	15	3	6	18
		Σfi = 30			Σfiui = – 3	Σ fiui2 = 75

Question 10.
Find the mean and the standard deviation from the following :

Wages (in ₹)	120-200	200-210	210-220	220-230
No. of workers	10	12	18	20
Wages (in ₹)	230-240	240-250	250-260	260-270
No. of workers	25	18	16	5

Solution:

Wages (in ₹)	No. of workers fi	xi	di – xi – A A = 225	di2	fidi	fidi2
120-200	10	160	-65	4225	-650	42250
200-210	12	205	-20	400	-240	4800
210-220	18	215	-10	100	-180	1800
220 – 230	20	225	‘ 0	0	0	0
230-240	25	235	10	100	250	2500
240-250	18	245	20	400	360	7200
250 – 260	16	255	30	900	480	14400
260-270	5	265	40	1600	200	8000
	Σfi = 124				Σ fidi = 220	Σ fidi2 = 80950

Question 11.
The following table shows the I.Q. of 480 school children. Find
(i) the mean.
(ii) the standard deviation using the step deviation method. Use Chrlier’s check to verify the computation of the standard deviation.

X	70	74	78	82	86	90	94	98	102	106	110	114	118	122	126
f	4	9	16	28	45	66	85	72	54	38	27	18	11	5	2

Solution:
The table of values is given as under :

xi	fi	di = xi – A A = 98	\(\frac{d_i}{i}\) = ui I = 4	ui2	fiui	fiui2
70	4	-28	-7	49	-28	196
74	9	-24	-6	36	-54	324
78	16	-20	-5	25	-80	400
82	28	– 16	-4	16	– 112	448
86	45	– 12	-3	9	-135	405
90	66	-8	-2	4	– 132	264
94	85	-4	-1	1	-85	85
98	72	0	0	0	0	0
102	54	4	1	1	54	54
106	38	8	2	4	76	152
110	27	12	3	9	81	243
114	18	16	4	16	72	288
118	11	20	5	25	55	275
122	5	24	6	36	30	180
126	2	28	7	49	14	98
	Σfi = 480				Σfiui = -244	Σ fiui2 = 3412

Question 12.
In a certain test, the 30 scores were grouped as follows :

30-34	35-39	40-44	45-49	50-54	55-59	60-64
2	2	7	10	6	2	1

Calculate the mean and the standard deviation:
Solution:
The table of values is given as under:

Intervals	frequency	xi	di = xi – 47	ui = \(\frac{d_i}{5}\)	fiui	fiui2
30-34	2	32	-15	-3	-6	18
35-39	2	37	– 10	-2	-4	8
40-44	7	42	-5	-1	-7	7
45-49	10	47	0	0	0	0
50-54	6	52	5	1	6	6
55-59	2	57	10	2	4	8
60-64	1	62	15	3	3	9
	Σfi = 30				Σfiui = – 4	Σfiui2 = 56

Then by step deviation method, we have

Question 13.
The number of faults on the surface of each of 1000 tiles were distributed as follows :

No. of faults	0	1	2	3	4	5
Frequency	760	138	67	25	8	2

Calculate the mean and the standard deviation.
Solution:
The table of values is given as under :

xi	fi	fixi	fixi2
0	760	0	0
1	138	138	138
2	67	134	268
3	25	75	225
4	8	32	128
5	2	10	50
	Σfi = 1000	Σ fixi = 389	Σ fixi2 = 809

Question 14.
The mean and the standard deviation of 25 observations and 60 and 3. Later on it was decided to omit an observation which was incorrectly recorded as 50. Calculate the mean and the standard deviation of the remaining 24 observations.
Solution:

Question 15.
The scores of two golfers for 10 rounds each are:

A	58	59	60	54	65	66	52	75	69	52
B	84	56	92	65	86	78	44	54	78	68

Which may be regarded as the more consistent player ?
Solution:
For golfer A ; no. of observations = 10
and Sum of all observations = 58 + 59 + 60 + 54 + 65 + 66 + 52 + 75 + 69 + 52 = 610
∴Mean \(\overline{\mathrm{x}}\) = \(\frac { Sum of all observations }{ 10 }\) = \(\frac { 610 }{ 10 }\) = 61
For golfer B : Sum of all observations = 84 + 56 + 92 + 65 + 86 + 78 + 44 + 54 + 78 + 68 = 705
∴Mean \(\overline{\mathrm{Y}}\) = \(\frac { Sum of all observations }{ 10 }\) = \(\frac { 705 }{ 10 }\) = 70.5
We construct the table of values is given as under :

X	Y	(X-\(\bar{X}\))	(X-\(\bar{X}\))2	Y-\(\bar{Y}\)	(Y-\(\bar{Y}\))2
58	84	-3	9	13.5	182.25
59	56	-2	4	– 14.5	210.25
60	92	– 1	1	21.5	462.25
54	65	-7	49	-5.5	30.25
65	86	4	16	15.5	240.25
66	78	5	25	7.5	56.25
52	44	-9	81	-26.5	702.25
75	54	14	196	-16.5	272.25
69	78	8	64	7.5	56.25
52	68	-9	81	-2.5	6.25
			I (X-\(\bar{X}\))2 =526		E(X-\(\bar{X}\))2 =2218.50

Question 16.
Goals scored by two teams A and B in a football season were as follows :

Number of goods scored in a match	Number of Matches
	A	B
0	27	17
1	9	9
2	8	6
3	5	5
4	4	3

By calculating the coefficient of variation in each case find which team may be considered more consistent.
Solution:
The table of values is given as under :

xi	fA	fB	fAxi	fBxi	fA xi2	fB xi2
0	27	17	0	0	0	0
1	9	9	9	9	9	9
2	8	6	16	12	32	24
3	5	5	15	15	45	45
4	4	31	16	12	64	48
	Σ fA = 53	Σ fB = 40	Σ fAxi = 56	Σ fBxi = 48	Σ fA xi2 = 150	Σ fB xi2 = 126

Question 17.
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b ?
(a) a = 0, b = 7
(b) a = 5, b = 2
(c) a = l, b = 6
(d) a = 3, b = 4
Solution:
Given observations are a, b, 8, 5 and 10 ∴ no. of observations = n = 5, \(\overline{\mathrm{x}}\) = 6, \(\sigma^2\) = 6.80
∴ Mean = \(\frac { Sum of all given observations }{ n }\) ⇒ 6 = \(\frac { a+b+8+5+10 }{ 5 }\)
⇒ 30 = a + b + 23
⇒ a + b = 7

From (1) and (2); we have
a² + (7 – a)² = 25
⇒ 2a² – 10a + 24 = 0
⇒ a² – 7a + 12 = 0
⇒ (a – 3)(a – 4) = 0
⇒ a = 3, 4
When a = 3 ∴ from (1); b = 4
When a = 4 ∴ from (1); b = 3
∴ Ans. (d)