RD Sharma Solutions Class 10 Statistics Exercise 7.3

RD Sharma Class 10 Solutions Chapter 7 Ex 7.3 PDF Free Download

Exercise 7.3

 

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 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

From the above table, we can conclude that,

A = 275, h = 50

Mean = A + h x sum/N

= 275 + 50 x -35/200

= 275 – 8.75 = 266.25

Therefore, mean expenditure (in rupees) per household= 266.25

 

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

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

Solution:

We know that,

class marks (x) = (upper class limit + lower class limit)/2

To find xi and fixi.

From the table given below:

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

We find that,

N = 20

Sum = 162

Mean \(\overline{x}\) = Sum/N

162/20 = 8.1

Average number of plants per house =  8.1

Therefore, mean number of plants per house= 8.1

 

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 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

From the above table, we can conclude that,

N = 50, h = 20

Mean = A + h x sum/N

= 150 + 2 x (-15)/5

= 150 – 4.8

= 145.2

Therefore, mean daily wages of the workers of the factory= 145.2

 

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:

Marks of each interval (xi) = (upper class limit + lower class limit)/2

Class size of this data = 3

Let assumed mean (a)= 75.5

We may calculate di, ui, fiui from the following table,

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

From the above table, we conclude that,

N = 30, sum = 4

Mean \(\overline{x}\) = 75.5 + (4/3)x3

= 75.5 + 0.4

= 75.9

Therefore, the mean heart beats per minute for those women =75.9 beats per minute

 

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 the mean =15

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

From the above table,

A = 15, h = 6

Mean = A + h(sum/A)

= 15 + 6(3/40)

= 15 + 0.45

= 15.45

Therefore, the mean= 15.45

 

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

6.

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

Solution:

Let the mean = 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
61

From the above table, we can conclude that,

A = 100, h = 20

Mean    = 100 + 20 (61/100)

= 100 + 12.2

= 112.2

Therefore, the mean=112.2

 

7.

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

Solution:

Let the mean (A) = 20

Class interval Mid- value di= xi – 20 ui = (xi – 20)/8 fi fiui
0-8 4 -16 -2 6 -12
8-16 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

From the above table, we can conclude that,

A = 20, h = 8

Mean= A + h (sum/N)

= 20 + 8 (7/40)

= 20 + 1.4

= 21.4

Therefore, the mean= 21.4

 

8.

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

Solution:

Let the mean (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

From the above table, we can conclude that,

A = 15, h = 6

Mean = A + h(sum/N)

= 15 + 6 (5/40)

= 15 + 0.75

= 15.75

Therefore, the mean= 15.75

 

9.

Class interval: 0 – 10                    10 – 20                  20 – 30                  30 – 40                  40 – 50
Frequency: 9                              12                           15                           10                           14

Solution:

Let the mean (A) = 25

Class interval Mid – value di = xi -25 ui = (xi -25)/10 Frequency fi fiui
0 – 10 5 -20 -2 9 -18
10 – 20 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

From the above table, we can conclude that,

A = 25, h = 10

Mean    = A + h(sum/N)

= 25 + 19 (8/60)

= 25 + (4/3)

= 26.333

Therefore, the mean= 26.333

10.

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

Solution:

Let the 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

From the above table, we can conclude that,

A = 20, h = 8

Mean    = A + h (sum/N)

= 20 + 8 (5/ 40)

= 20 + 1

= 21

Therefore, the mean=21

 

11.

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

Solution:

Let the 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

From the above table, we can conclude that,

A = 20, h = 8

Mean = A + h (sum/N)

= 20 + 8 (-9/ 20)

= 20 – (72/20)

= 20 – 3.6

= 16.4

Therefore, the mean= 16.4

 

12.

Class interval: 10-30     30-50     50-70     70-90     90-110   110-130
Frequency: 5              8              12           20           3              2

Solution:

Let the 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

From the above table, we can conclude that,

A = 60, h = 20

Mean = A + h (sum/N)

= 60 + 20 (14/ 5)

= 60 + 5.6

= 65.6

Therefore, the mean=65.6

 

13.

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

Solution:

Let the 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

From the above table, we can conclude that,

A = 50, h = 10

Mean    = A + h (sum/N)

= 50 + 10 (-2/ 40)

= 50 – 0.5

= 49.5

Therefore, the mean=49.5

 

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 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

From the above table, we can conclude that,

A = 42, h = 5

Mean = A + h (sum/N)

= 42 + 5 (-79/70)

= 42 – 79/14

= 36.357

Therefore, the mean= 36.357

 

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:

Using direct method,

Class interval Mid value xi Frequency fi fixi
1 – 4 2.5 6 15
4 – 9 6.5 12 18
9 – 16 12.5 26 325
16 – 27 21.5 20 430
N = 64 Sum = 848

From the above table, we can conclude that,

Mean = (sum/N) + A

= 848/ 64

= 13.25

Using the mean method

Let the 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

From the above table, we can conclude that,

Mean = A + sum/N

= 6.5 + 6.75

= 13.25

Therefore, the mean= 13.25

 

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 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

From the above table, we can conclude that,

A = 16, h = 100

Mean = A + h (sum/N)

= 1650 + 100 (7/52)

= 1650 + (175/13)

= 21625/13

= 1663.46

Therefore, the weekly cost of living index= 1663.46

 

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

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

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

From the above table, we can conclude that,

Mean    = sum/ N

= 3620/ 140

= 25.857

Therefore, the mean= 25.857

(ii)Assumed mean method:

Let the 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

From the above table, we can conclude that,

Mean = A + (sum/ N)

= 25 + (120/ 140)

= 25 + 0.857

= 25.857

Therefore, the mean=25.857

(iii)Step deviation method:

Let the 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

From the above table, we can conclude that,

Mean = A + h(sum/ N)

= 25 + 10(12/ 140)

= 25 + 0.857

= 25.857

Therefore, the mean= 25.857

 

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 = 50 Sum = 30f1+70f2+ 2060

From the above table, we can conclude that,

Sum of frequency = 50

5 + f1 + 10 + f2 + 7 + 8 = 50

f1 + f2 = 20

3f1 + 3f2 = 60 —- (i) [multiply both side by 3]

And mean = 62.8

Sum/ N = 62.8

(30f1+70f2+ 2060)/ 50 = 62.8

30f1+70f2 = 3140 – 2060

30f1+70f2 = 1080

3f1 + 7f2 = 108 —- (ii) [divide it by 10]

subtract equation (i) from equation (ii)

3f1 + 7f2 – 3f1 – 3f2 = 108 – 60

4f2 = 48

f2 = 12

Substituting the value of f2 in equation (i)

3f1 + 3(12) = 60

f1 = 24/3 = 8

f1 = 8, f2 = 12

Therefore, f1 = 8, f2 = 12

 

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

From the above table, we can conclude that,

Mean = sum/ N

18 = \(\frac {752+ 20x}{44+ x}\)

792 + 18x = 752 + 20x

2x = 40

x = 20

Therefore, the missing frequency= 20

 

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

From the above table, we can conclude that,

Given mean =27

Mean = sum/ N

\(\frac {1245 +15p}{43+ p}\) = 27

1245 + 15p = 1161 + 27p

12p = 84

P = 7

Therefore, the value of P= 7

 

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

From the above table, we can conclude that the class internals are not continuous

Since there is a gap between two class intervals.

So we have to add ½  from lower class limit of each interval and class mark (xi) may be obtained by using the relation

xi = \(\frac {upper limit + lower class limit}{2}\)

Class size (h) of this data = 3

Now taking 57 as mean (a) we may calculate di ,ui, fiui as follows

Class interval Frequency fi Mid value xi

 

di= xi – A

= xi ­– 25

ui = (xi – 25)

10

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

From the above table, we can conclude that,

N = 400

Sum = 25

Mean = A + h (sum/ N)

= 57 + 3 (45/400)

= 57 + 3/ 16

= 57+ 0.1875

= 57.19

Therefore, mean number of mangoes kept in packing box =57.19

 

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 = \(\frac {upper limit + lower class limit}{2}\)

Class size = 50

Now, talking 225 as 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

From the above table, we can conclude that,

N = 25

Sum = -7

Mean \(\overline{x}=a+\left( \frac{sum}{N}\right)\times h\)

225 + 50 (-7/ 25)

225 – 14 = 211

Therefore, mean daily expenditure on food =Rs 211

 

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

\(x=\frac {upper limit + lower class limit}{2}\)

Class size of this data = 0.04

Now taking 0.04 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 0.10 -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 above table, we can conclude that,

N = 30

Sum = -31

Mean \(\overline{x}=a+\left( \frac{sum}{N}\right)\times h\)

= 0.14 +(0.04)(-31/30)

= 0.099 ppm

Therefore, mean concentration of SO2 ­in the air is 0.099 ppm

 

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

\(x=\frac {upper limit + lower class limit}{2}\)

Now, taking 16 as mean (a) we may

Calculate di and fidi  as follows

Number of days Number of students fi Xi d = xi + 10 fidi
0 – 6 11 3 -13 -143
6 – 10 10 8 -8 -280
10 – 14 7 12 -4 -28
14 – 20 7 16 0 0
20 – 28 8 24 8 32
28 – 36 3 33 17 51
30 – 40 1 39 23 23
Total N = 40 Sum = -145

From the above table, we can conclude that,

N = 40

Sum= -145

Mean \(\overline{x}=a+\left( \frac{sum}{N}\right) \)

= 16 + (-145/ 40)

= 16 – 3.625

= 12.38

Therefore, mean number of days  for which student was absent =12.38 days,

 

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

\(x=\frac {upper limit + lower class limit}{2}\)

Class size (h) for this data = 10

Now taking 70 as assumed mean (a)

Calculate di ,ui, fiui as follows

Litracy rate (in %) Number of cities (fi) Mid value xi

 

di= xi ­– 70 ui di

50

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

From the above table, we can conclude that,

N = 35

Sum = -2

Mean \(\overline{x}=a+\left( \frac{sum}{N}\right)\times h\)

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

Therefore, mean literacy rate = 69.43 %

 

Leave a Comment

Your email address will not be published. Required fields are marked *