Chapter 7: Statistics Exercise – 7.3

Question: 1

The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.

Expenditure (in rupees) (x)	Frequency (fi)	Expenditure (in rupees) (xi)	Frequency (fi)
100 – 150	24	300 – 350	30
150 – 200	40	350 – 400	22
200 – 250	33	400 – 450	16
250 – 300	28	450 – 500	7

Find the average expenditure (in rupees) per household

Solution:

Let the assumed mean (A) = 275

Class interval	Mid value (xi)	di = xi – 275	ui = (xi - 275)/50	Frequency fi	fiui
100 – 150	125	-150	-3	24	-12
150 – 200	175	-100	-2	40	-80
200 – 250	225	-50	-1	33	-33
250 – 300	275	0	0	28	0
300 – 350	325	50	1	30	30
350 – 400	375	100	2	22	44
400 – 450	425	150	3	16	48
450 – 500	475	200	4	7	28
				N = 200	Sum = -35

We have A = 275, h = 50
Mean = A + h * sum/N
= 275 + 50 * - 35/200
= 275 – 8.75
= 266.25

 

Question: 2

A survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.

Number of Number of plants:	0 - 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
Number of houses:		1	2	5	6	2	3

Which method did you use for finding the mean, and why?

Solution:

Let us find class marks (x_i­) = (upper class limit + lower class limit)/2. Now we may compute x_i and f_ix_i as following.

Number of plants	Number of house (fi)	xi	Fixi
0 - 2	1	1	1
2 – 4	2	3	6
4 – 6	1	5	5
6 – 8	5	7	35
8 – 10	6	9	54
10 – 12	2	11	22
12 – 14	3	13	39
Total	N = 20		Sum = 162

From the table we may observe that N = 20, Sum = 162

So mean number of plants per house is 8.1. We have used for the direct method values Xi and fi are very small

 

Question: 3

Consider the following distribution of daily wages of workers of a factory

Daily wages (in Rs)	100 - 120	120 - 140	140 - 160	160 - 180	180 - 200
Number of workers:	12	16	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

Let the assume mean (A) = 150

Class interval	Mid value xi	di = xi - 150	ui = (xi - 150)/20	Frequency fi	Fiui
100 – 120	110	-40	-2	12	-24
120 – 140	130	-20	-1	14	-14
140 – 160	150	0	0	8	0
160 – 180	170	20	1	6	6
180 – 200	190	40	2	10	20
				N = 50	Sum = -12

We have N = 50, h = 20

= 150 - 4.8
= 145.2

 

Question: 4

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method. Number of heart

Beats Per minute:	65 - 68	68 - 71	71 - 74	74 - 77	77 - 80	80 - 83	83 - 86
Number of women:	2	4	3	8	7	4	2

Solution:

We may find marks of each interval (xi) by using the relation (xi¬) = (upper class limit + lower class limit)/2 Class size of this data = 3 Now talking 75.5 as assumed mean (a) We may calculate di, ui, fiui as following

Number of heart beats per minute	Number of women (xi)	xi	di = xi – 75.5	ui = (xi - 755)/h	fiui
65-68	2	66.5	-9	-3	-6
68-71	9	69.5	-6	-2	-8
71-74	3	72.5	-3	-1	-3
74-77	8	75.5	0	0	0
77-80	7	78.5	3	1	7
80-83	4	81.5	6	2	8
83-86	2	84.5	9	3	6
	N = 30				Sum = 4

Now we may observe from table that N = 30, sum = 4

= 75.5 + 0.4
= 75.9 
So mean heart beats per minute for those women are 75.9 beats per minute

 

Question: 5

Find the mean of each of the following frequency distributions: (5 - 14)

Class interval:	0 - 6	6 – 12	12 – 18	18 – 24	24 – 30
Frequency:	6	8	10	9	7

Solution:

Let us assume mean be 15

Class interval	Mid - value	di = xi – 15	ui = (xi – 15)/6	fi	fiui
0 - 6	3	-12	-2	6	-12
12-Jun	9	-6	-1	8	-8
18-Dec	15	0	0	10	0
18 - 24	21	6	1	9	9
24 - 30	27	18	2	7	14
				N = 40	Sum = 3

A = 15, h = 6, N = 40

= 15 + 0.45

= 15.47

 

Question: 6

Class interval:	50 - 70	70 - 90	90 - 110	110 - 130	130 - 150	150 - 170
Frequency:	18	12	13	27	8	22

Solution:

Let us assumed mean be 100

Class interval	Mid-value xi	di = xi – 100	ui = (xi – 100)/20	fi	fiui
50 - 70	60	-40	-2	18	-36
70 – 90	80	-20	-1	12	-12
90 – 110	100	0	0	13	0
110 – 130	120	20	1	27	27
130 – 150	140	40	2	8	16
150 – 170	160	60	3	22	66
				N = 100	Sum = 61

A = 100, h = 20

= 100 + 12.2
= 122.2

 

Question: 7

Class interval:	0- 8	8 – 16	16 - 24	24 - 32	32 - 40
Frequency:	6	7	10	8	9

Solution:

Let the assumed mean (A) = 20

Class interval	Mid- value	di= xi - 20	ui = (xi – 20)/8	fi	fiui
0-8	4	-16	-2	6	-12
16-Aug	12	-8	-1	7	-7
16-24	20	0	0	10	0
24-32	28	8	1	8	8
32-40	36	16	2	9	18
				N = 40	Sum = 7

We have A = 20, h = 8 
Mean = A + h (sum/N) 
= 20 + 8 (7/40) 
= 20 + 1.4 
= 21.4 

 

Question: 8

Class interval:	0 - 6	6 – 12	12 – 18	18 - 24	24 - 30
Frequency:	7	5	10	12	6

Solution:

Let the assumed mean be (A) = 15

Class interval	Mid – value	di = xi -15	ui = (xi -15)/6	Frequency fi	fiui
0 – 6	3	-12	-2	-1	-14
6 – 12	9	-6	-1	5	-5
12 – 18	15	0	0	10	0
18 – 24	21	6	1	12	12
24 – 30	27	12	2	6	12
				N = 40	Sum = 5

We have A = 15, h = 6 
Mean = A + h(sum/N) 
= 15 + 6 (5/40) 
= 15 + 0.75 
= 15.75

 

Question: 9

Class interval:	0 - 10	20 – 30	20 - 30	30 - 40	40 - 50
Frequency:	6	7	10	8	9

Solution:

Let the assumed mean (A) = 25

Class interval	Mid – value	di = xi -25	ui = (xi - 25)/10	Frequency fi	fiui
0 - 10	5	-20	-2	9	-18
20-Oct	15	-10	-1	10	-12
20 - 30	25	0	0	15	0
30 - 40	35	10	1	10	10
40 - 50	45	20	2	14	28
				N = 60	Sum = 8

We have A = 25, h = 10 
Mean = A + h (sum/N) 
= 25 + 19 (8/60) 
= 25 + (4/3) 
= 26.333 

 

Question: 10

Class interval:	0- 8	8 – 16	16 - 24	24 - 32	32 - 40
Frequency:	5	9	10	8	8

Solution:

Let the assumed mean (A) = 20

Class interval	Mid value xi	di= xi - 20	ui = (xi -20)/8	Frequency fi	fiui
0 – 8	4	-16	-2	5	-10
8 – 16	12	-8	-1	9	-9
16 – 24	20	0	0	10	0
24 – 32	28	8	1	8	8
32 – 40	36	16	2	8	16
				N = 40	Sum = 5

We have, A = 20, h = 8 
Mean = A + h (sum/N) 
= 20 + 8 (5/ 40) 
= 20 + 1
= 21

 

Question: 11

Class interval:	0- 8	8 – 16	16 - 24	24 - 32	32 - 40
Frequency:	5	6	4	3	2

Solution:

Let the assumed mean (A) = 20

Class interval	Mid value xi	di= xi - 20	ui = (xi -20)/8	Frequency fi	fiui
0 – 8	4	-16	-2	-2	-10
8  – 16	12	-8	-1	-1	-6
16 – 24	20	0	0	0	0
24 – 32	28	8	1	1	3
32 – 40	36	16	2	2	4
				N = 20	Sum = -9

We have, A = 20, h = 8 
Mean = A + h (sum/N) 
= 20 + 8 (-9/ 20) 
= 20 – (72/20) 
= 20 – 3.6 
= 16.4 

 

Question: 12

Class interval:	20 – 30	30 - 50	50 - 70	70 - 90	90- 110	110 - 130
Frequency:	5	8	12	20	3	2

Solution:

Let the assumed mean (A) = 60

Class interval	Mid value xi	di= xi - 60	ui = (xi -60)/20	Frequency fi	fiui
10 – 30	20	-40	-2	5	-10
30 – 50	40	-20	-1	8	-8
50 – 70	60	0	0	12	0
70 – 90	80	20	1	20	20
90 – 110	100	40	2	3	6
110 – 130	120	60	3	2	6
				N = 50	Sum = 14

We have A = 60, h = 20 
Mean = A + h (sum/N) 
= 60 + 20 (14/ 5) 
= 60 + 5.6 
= 65.6

 

Question: 13

Class interval:	25 - 35	35 – 45	45 - 55	55 - 65	65 - 75
Frequency:	6	10	8	10	4

Solution:

Let the assumed mean (A) = 50

Class interval	Mid value xi	di= xi - 50	ui = (xi - 50)/ 10	Frequency fi	fiui
25 – 35	30	-20	-2	6	-12
35 – 45	40	-10	-1	10	-10
45 – 55	50	0	0	8	0
55 – 65	60	10	1	12	12
65 – 75	70	20	2	4	8
				N = 40	Sum = -2

We have A = 50, h = 10 
Mean = A + h (sum/N) 
= 50 + 10 (-2/ 40) 
= 50 - 0.5 
= 49.5 

 

Question: 14

Class interval:	25 - 29	30 - 34	35 -39	40 - 44	45 - 49	50 - 54	55 - 59
Frequency:	14	22	16	6	5	3	4

Solution:

Let the assumed mean (A) = 42

Class interval	Mid value xi	di = xi - 42	ui = (xi - 42)/ 5	Frequency fi	fiui
25 – 29	27	-15	-3	14	-42
30 – 34	32	-10	-2	22	-44
35 – 39	37	-5	-1	16	-16
40 – 44	42	0	0	6	0
45 – 49	47	5	1	5	5
50 – 54	52	10	2	3	6
55 – 59	57	15	3	4	12
				N = 70	Sum = -79

We have A = 42, h = 5 
Mean = A + h (sum/N) 
= 42 + 5 (-79/70) 
= 42 – 79/14 
= 36.357

 

Question: 15

For the following distribution, calculate mean using all suitable methods:

Size of item:	1 – 4	4 – 9	9 – 16	16 -20
Frequency:	6	12	26	20

Solution:

By direct method

Mean = (sum/N) 
= 848/ 64= 13.25 

By assuming mean method Let the assumed mean (A) = 65

Class interval	Mid value xi	ui = (xi - A) = xi - 65	Frequency fi	fiui
1 – 4	2.5	-4	6	-25
4 – 9	6.5	0	12	0
9 – 16	12.5	6	26	196
16 – 27	21.5	15	20	300
			N = 64	Sum = 432

Mean = A + sum/N 
= 6.5 + 6.75 
= 13.25

 

Question: 16

The weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost of living index.

Cost of living index	Number of students	Cost of living index	Number of students
1400 – 1500	5	1700 – 1800	9
1500 – 1600	10	1800 – 1900	6
1600 – 1700	20	1900 – 2000	2

Solution:

Let the assumed mean (A) = 1650

Class interval	Mid value xi	di = xi – A = xi– 1650	ui = (xi – 1650)100	Frequency fi	fiui
1400 – 1500	1450	-200	-2	5	-10
1500 – 1600	1550	-100	-1	10	-10
1600 – 1700	1650	0	0	20	0
1700 – 1800	1750	100	1	9	9
1800 – 1900	1850	200	2	6	12
1900 – 2000	1950	300	3	2	6
				N = 52	Sum = 7

We have A = 16, h = 100 
Mean = A + h (sum/N)
= 1650 + 100 (7/52)
= 1650 + (175/13) 
= 21625/13 
= 1663.46

 

Question: 17

The following table shows the marks scored by 140 students in an examination of a certain paper:

Marks:	0 - 10	10 – 20	20 - 30	30 - 40	40 - 50
Number of students:	20	24	40	36	20

Solution:

(i) Direct method:

Class interval	Mid value xi	Frequency fi	fixi
0 – 10	5	20	100
10 – 20	15	24	360
20 – 30	25	40	1000
30 – 40	35	36	1260
40 – 50	45	20	900
		N = 140	Sum = 3620

Mean = sum/ N 
= 3620/ 140 
= 25.857

(ii) Assumed mean method: Let the assumed mean = 25 Mean = A + (sum/ N)

Class interval	Mid value xi	ui = (xi - A)	Frequency fi	fiui
0 – 10	5	-20	20	-400
10 – 20	15	-10	24	-240
20 – 30	25	0	40	0
30 – 40	35	10	36	360
40 – 50	45	20	20	400
			N = 140	Sum = 120

Mean = A + (sum/ N)
= 25 + (120/ 140)
= 25 + 0.857
= 25.857

(iii) Step deviation method: Let the assumed mean (A) = 25

Class interval	Mid value xi	di= xi – A = xi– 25	ui = (xi – 25)10	Frequency fi	fiui
0 – 10	5	-20	-2	20	-40
10 – 20	15	-10	-1	24	-24
20 – 30	25	0	0	40	0
30 – 40	35	10	1	36	36
40 – 50	45	20	2	20	40
				N = 140	Sum = 12

Mean = A + h (sum/ N)
= 25 + 10(12/140)
= 25 + 0.857
= 25.857

 

Question: 18

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the miss frequency f₁ and f₂.

Class:	0 - 20	20 – 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency:	5	f1	10	f2	7	8

Solution:

Class interval	Mid value xi	Frequency fi	fixi
0 – 20	10	5	50
20 – 40	30	f1	30f1
40 – 60	50	10	500
60 – 80	70	f2	70f2
80 – 100	90	7	630
100 – 120	110	8	880
		N = 30 +fi + f2	Sum = 30f1+70 f2+ 2060

Given, sum of frequency
50 = f₁ + f₂ + 30
f₁ + f₂ = 50 – 30
f₁ +f₂ = 20
3f₁ + 3f₂ = 60 ---- (1)       [multiply both side by 3] 
And mean = 62.8 Sum/ N 
= 62.8 = (30f₁+ 70f₂+ 2060)/50 
= 3140 = 30f₁ + 70f₂ + 2060
30f₁+ 70f₂ = 3140 – 2060
30f₁ + 70f₂ = 1080
3f₁ + 7f₂ = 108 ---- (2) [divide it by 10]
Subtract equation (1) from equation (2)
3f₁ + 7f₂ - 3f₁ - 3f₂ = 108 – 60
4f₂ = 48 = f₂ = 12

Put value of f₂ in equation (1)
3f₁ + 3(12) = 60
f₁ = 24/3 = 8
 f₁ = 8, f₂ = 12

 

Question: 19

The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs 18.00. Find out the missing frequency.

Class interval:	11 – 13	13- 15	15 - 17	17 - 19	19 - 21	21 - 23	23 - 25
Frequency:	7	6	9	13	-	5	4

Solution:

Given mean = 18, Let the missing frequency be v

Class interval	Mid value xi	Frequency fi	fixi
11 – 13	12	7	84
13 – 15	14	6	88
15 – 17	16	9	144
17 – 19	18	13	234
19 – 21	20	x	20x
21 – 23	22	5	110
23 – 25	14	4	56
		N = 44 + x	Sum = 752 + 20x

792 + 18x = 752 + 20x
20x – 18x = 792 – 752
x = 40/2
x = 20

 

Question: 20

If the mean of the following distribution is 27. Find the value of p.

Class:	0 -10	10 – 20	20 - 30	30 - 40	40 - 50
Frequency:	8	p	12	13	10

Solution:

Class interval	Mid value xi	Frequency fi	fixi
0 – 10	5	8	40
10 – 20	15	P	152
20 – 30	25	12	300
30 – 40	35	13	455
40 – 50	45	16	450
		N = 43 + P	Sum = 1245 + 15p

Given Mean = 27
Mean = sum/ N

1161 + 27p = 1245 + 15p
27p – 15p = 1245 – 1161
12p = 84
p = 84/12
p = 7 

 

Question: 21

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes:	50 - 52	53 - 55	56 - 58	59 - 61	62 - 64
Number of boxes:	15	110	135	115	25

Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?

Solution:

Number of mangoes	Number of boxes (fi)
50 – 52	15
53 – 55	110
56 – 58	135
59 – 61	115
62 – 64	25

We may observe that class internals are not continuous There is a gap between two class intervals. So we have to add 1/2 from lower class limit of each interval and class mark (xi) may be obtained by using the relation xi = (upper limit + lower class limit)/2. Class size (h) of this data = 3 Now taking 57 as assumed mean (a) we may calculated di , ui, fiui as follows

Class interval	Frequency fi	Mid value xi	di =  xi - A	ui = (xi – A)/3	fiui
49.5 – 52.5	15	51	-6	-2	-30
52.5 – 55.5	110	54	-3	-1	-110
55.5 – 58.5	135	57	0	0	0
58.5 – 61.5	115	60	3	1	115
61.5 – 64.5	25	63	6	2	50
Total	N = 400				Sum = 25

Now we have N = 400, Sum = 25, 
Mean = A + h (sum/ N) 
= 57 + 3 (25/400) 
= 57 + 3/16 
= 57+ 0.1875 
= 57.19
Clearly mean number of mangoes kept in packing box is 57.19 

 

Question: 22

The table below shows the daily expenditure on food of 25 households in a locality

Daily expenditure (in Rs):	100 - 150	150 - 200	200 - 250	250 -300	300 - 350
Number of households:	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Solution:

We may calculate class mark (xi) for each interval by using the relation xi = (upper limit + lower class limit)/2. Class size = 50 Now, talking 225 as assumed mean (xi) we may calculate di ,ui, fiui as follows:

Daily expenditure	Frequency fi	Mid value xi	di = xi ­– 225	ui = (xi – 225)50	fiui
100 – 150	4	125	-100	-2	-8
150 – 200	5	175	-50	-1	-5
200 – 250	12	225	0	0	0
250 – 300	2	275	50	1	2
300 – 350	2	325	100	2	4
	N = 25				Sum = – 7

Now we may observe that N = 25 Sum = -7

225 = 50 + (-7/ 25) × 225 
225 – 14 
= 211 
So, mean daily expenditure on food is Rs 211

 

Question: 23

To find out the concentration of SO2 in the air (in parts per million i. e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:

Concentration of SO2 (in ppm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	9
0.08 – 0.12	9
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

Find the mean concentration of SO₂ in the air 

Solution:

We may find class marks for each interval by using the relation xi = (upper limit + lower class limit)/2. Class size of this data = 0.04 Now taking 0.04 assumed mean (xi) we may calculate di, ui, fiui as follows:

Concentration of SO2	Frequency fi	Class interval xi	di = xi – 0.14	ui	fiui
0.00 – 0.04	4	0.02	-0.12	-3	-12
0.04 – 0.08	9	0.06	-0.08	-2	-18
0.08 – 0.12	9	0.1	-0.04	-1	-9
0.12 – 0.16	2	0.14	0	0	0
0.16 – 0.20	4	0.18	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	N = 30				Sum = -31

From the table we may observe that N = 30, Sum = - 31

= 0.04 + (-31/30) × (0.04) 
= 0.099 ppm 
So mean concentration of SO2 ¬in the air is 0.099 ppm 

 

Question: 24

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days:	0 - 6	6 – 10	10 – 14	14 - 20	20 - 28	28 - 38	38 - 40
Number of students:	11	10	7	4	4	3	1

Solution:

We may find class mark of each interval by using the relation xi = (upper limit + lower class limit)/2 Now, taking 16 as assumed mean (a) we may Calculate di and fi di  as follows

Number of days	Number of students fi	Xi	d = xi - 16	fidi
0 – 6	11	3	-13	-143
6 – 10	10	8	-8	-80
10 – 14	7	12	-4	-28
14 – 20	7	16	0	0
20 – 28	8	24	8	64
28 – 36	3	33	17	51
30 – 40	1	39	23	23
Total	N = 40			Sum = -113

Now we may observe that N = 40, Sum = -113

16 + (-113/ 40)
= 16 – 2.825
= 13.75
So mean number of days is 13.75 days, for which student was absent

 

Question: 25

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %):	45 -55	55 - 65	65 - 75	75 - 85	85 - 95
Number of cites:	3	10	11	8	3

Solution:

We may find class marks by using the relation x_i = (upper limit + lower class limit)/2 Class size (h) for this data = 10 Now taking 70 as assumed mean (a) wrong Calculate d_i ,u_i, f_iu_i as follows

Literacy rate (in %)	Number of cities (fi)	Mid value xi	di= xi ­– 70	ui = (di –A)10	fiui
45 – 55	3	50	-20	-2	- 6
55 – 65	10	60	-10	-1	-10
65 – 75	11	70	0	0	0
75 – 85	8	80	10	1	8
85 – 95	3	90	20	2	6
Total	N = 35				Sum = -2

Now we may observe that N = 35, Sum = -2

= 70 + (-2/35) × 10 
= 70 – 4/7 
= 70 – 0.57 
= 69.43 
So, mean literacy rate is 69.43 %