 # 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) = (upper class limit + lower class limit)/2. Now we may compute xi and fixi 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 fand 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 SOin 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 xi = (upper limit + lower class limit)/2 Class size (h) for this data = 10 Now taking 70 as assumed mean (a) wrong Calculate d,ui, fiui 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 % ### Course Features

• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution