RD Sharma Class 10 Ex 7.3 Solutions Chapter 7 Statistics

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 7
Chapter NameStatistics
Exercise7.3
CategoryRD Sharma Solutions

RD Sharma Solutions for Class 10 Chapter 7 Statistics Ex 7.3 Download PDF

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 24300 – 35030
150 – 200 40350 – 40022
200 – 25033400 – 45016
250 – 30028450 – 5007

Find the average expenditure (in rupees) per household

Solution:

Let the assumed mean (A) = 275

Class intervalMid value (xi)d= xi – 275u= (x– 275)/50Frequency fifiui
100 – 150125-150-324-12
150 – 200175-100-240-80
200 – 250225-50-133-33
250 – 30027500280
300 – 3503255013030
350 – 40037510022244
400 – 45042515031648
450 – 5004752004728
    N = 200Sum = -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 – 22 – 44 – 66 – 88 – 1010 – 1212 – 14
Number of houses: 125623

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 plantsNumber of house (fi)xiFixi
0 – 2111
2 – 4236
4 – 6155
6 – 85735
8 – 106954
10 – 1221122
12 – 1431339
TotalN = 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 – 120120 – 140140 – 160 160 – 180180 – 200
Number of workers:12168610

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

Solution:

Let the assume mean (A) = 150

Class intervalMid value xid= xi – 150ui = (x– 150)/20Frequency fiFiui
100 – 120110-40-212-24
120 – 140130-20-114-14
140 – 1601500080
160 – 18017020166
180 – 2001904021020
    N = 50Sum = -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 – 6868 – 7171 – 7474 – 7777 – 80 80 – 83 83 – 86
Number of women:2438742

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 minuteNumber of women (xi)xidi = xi – 75.5ui = (x– 755)/hfiui
65-68266.5-9-3-6
68-71969.5-6-2-8
71-74372.5-3-1-3
74-77875.5000
77-80778.5317
80-83481.5628
83-86284.5936
 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 – 66 – 1212 – 1818 – 2424 – 30
Frequency:681097

Solution:

Let us assume mean be 15

Class intervalMid – valued= x– 15u= (x– 15)/6fifiui
0 – 63-12-26-12
12-Jun9-6-18-8
18-Dec1500100
18 – 24216199
24 – 3027182714
    N = 40Sum = 3

A = 15, h = 6, N = 40

= 15 + 0.45

= 15.47

Question: 6

Class interval:50 – 7070 – 9090 – 110110 – 130130 – 150150 – 170
Frequency:18121327822

Solution:

Let us assumed mean be 100

Class intervalMid-value xid= xi – 100u= (xi – 100)/20fifiui
50 – 7060-40-218-36
70 – 9080-20-112-12
90 – 11010000130
110 – 1301202012727
130 – 150140402816
150 – 1701606032266
    N = 100Sum = 61

A = 100, h = 20

= 100 + 12.2
= 122.2

Question: 7

Class interval:0- 8 8 – 1616 – 2424 – 3232 – 40
Frequency:671089

Solution:

Let the assumed mean (A) = 20

Class intervalMid- valuedi= x– 20ui = (x– 20)/8fifiui
0-84-16-26-12
16-Aug12-8-17-7
16-242000100
24-32288188
32-4036162918
    N = 40Sum = 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 – 66 – 1212 – 1818 – 2424 – 30
Frequency:7510126

Solution:

Let the assumed mean be (A) = 15

Class intervalMid – valued= x-15u= (x-15)/6Frequency fifiui
0 – 63-12-2-1-14
6 – 129-6-15-5
12 – 181500100
18 – 2421611212
24 – 3027122612
    N = 40Sum = 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 – 1020 – 3020 – 3030 – 4040 – 50
Frequency:671089

Solution:

Let the assumed mean (A) = 25

Class intervalMid – valued= x-25u= (x– 25)/10Frequency fifiui
0 – 105-20-29-18
20-Oct15-10-110-12
20 – 302500150
30 – 40351011010
40 – 50452021428
    N = 60Sum = 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- 88 – 1616 – 2424 – 3232 – 40
Frequency:591088

Solution:

Let the assumed mean (A) = 20

Class intervalMid value xidi= xi – 20u= (x-20)/8Frequency fifiui
0 – 84-16-25-10
8 – 1612-8-19-9
16 – 242000100
24 – 32288188
32 – 4036162816
    N = 40Sum = 5

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

Question: 11

Class interval:0- 88 – 1616 – 2424 – 3232 – 40
Frequency:56432

Solution:

Let the assumed mean (A) = 20

Class intervalMid value xidi= xi – 20u= (x-20)/8Frequency fifiui
0 – 84-16-2-2-10
8  – 1612-8-1-1-6
16 – 24200000
24 – 32288113
32 – 403616224
    N = 20Sum = -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 – 3030 – 50 50 – 7070 – 9090- 110110 – 130
Frequency:58122032

Solution:

Let the assumed mean (A) = 60

Class intervalMid value xidi= xi – 60u= (x-60)/20Frequency fifiui
10 – 3020-40-25-10
30 – 5040-20-18-8
50 – 706000120
70 – 90802012020
90 – 11010040236
110 – 13012060326
    N = 50Sum = 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 – 3535 – 4545 – 5555 – 6565 – 75
Frequency:6108104

Solution:

Let the assumed mean (A) = 50

Class intervalMid value xidi= xi – 50u= (x– 50)/ 10Frequency fifiui
25 – 3530-20-26-12
35 – 4540-10-110-10
45 – 55500080
55 – 65601011212
65 – 757020248
    N = 40Sum = -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 – 2930 – 3435 -3940 – 4445 – 4950 – 5455 – 59
Frequency:1422166534

Solution:

Let the assumed mean (A) = 42

Class intervalMid value xid= xi – 42u= (x– 42)/ 5Frequency fifiui
25 – 2927-15-314-42
30 – 3432-10-222-44
35 – 3937-5-116-16
40 – 44420060
45 – 49475155
50 – 545210236
55 – 5957153412
    N = 70Sum = -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 – 44 – 99 – 1616 -20
Frequency:6122620

Solution:

By direct method

Mean = (sum/N) 
= 848/ 64= 13.25 

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

Class intervalMid value xiu= (x– A) = x– 65Frequency fifiui
1 – 42.5-46-25
4 – 96.50120
9 – 1612.5626196
16 – 2721.51520300
   N = 64Sum = 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 indexNumber of studentsCost of living indexNumber of students
1400 – 150051700 – 18009
1500 – 1600101800 – 19006
1600 – 1700201900 – 20002

Solution:

Let the assumed mean (A) = 1650

Class intervalMid value xid= xi – A = xi– 1650u= (x– 1650)100Frequency fifiui
1400 – 15001450-200-25-10
1500 – 16001550-100-110-10
1600 – 1700165000200
1700 – 18001750100199
1800 – 190018502002612
1900 – 20001950300326
    N = 52Sum = 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 – 1010 – 2020 – 3030 – 4040 – 50
Number of students:2024403620

Solution:

(i) Direct method:

Class intervalMid value xiFrequency fifixi
0 – 10520100
10 – 201524360
20 – 3025401000
30 – 4035361260
40 – 504520900
  N = 140Sum = 3620

Mean = sum/ N 
= 3620/ 140 
= 25.857

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

Class intervalMid value xiu= (x– A)Frequency fifiui
0 – 105-2020-400
10 – 2015-1024-240
20 – 30250400
30 – 40351036360
40 – 50452020400
   N = 140Sum = 120

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

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

Class intervalMid value xidi= xi – A = xi– 25u= (x– 25)10Frequency fifiui
0 – 105-20-220-40
10 – 2015-10-124-24
20 – 302500400
30 – 40351013636
40 – 50452022040
    N = 140Sum = 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 – 2020 – 4040 – 6060 – 8080 – 100100 – 120
Frequency:5f110f278

Solution:

Class intervalMid value xiFrequency fifixi
0 – 2010550
20 – 4030f130f1
40 – 605010500
60 – 8070f270f2
80 – 100907630
100 – 1201108880
  N = 30 +fi + f2Sum = 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 – 1313- 1515 – 1717 – 1919 – 2121 – 2323 – 25
Frequency:7691354

Solution:

Given mean = 18, Let the missing frequency be v

Class intervalMid value xiFrequency fifixi
11 – 1312784
13 – 1514688
15 – 17169144
17 – 191813234
19 – 2120x20x
21 – 23225110
23 – 2514456
  N = 44 + xSum = 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 – 2020 – 3030 – 4040 – 50
Frequency:8p121310

Solution:

Class intervalMid value xiFrequency fifixi
0 – 105840
10 – 2015P152
20 – 302512300
30 – 403513455
40 – 504516450
  N = 43 + PSum = 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 – 5253 – 5556 – 5859 – 6162 – 64
Number of boxes:1511013511525

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

Solution:

Number of mangoesNumber of boxes (fi)
50 – 5215
53 – 55110
56 – 58135
59 – 61115
62 – 6425

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 intervalFrequency fiMid value xi  d=  xi – Au= (x– A)/3fiui
49.5 – 52.51551-6-2-30
52.5 – 55.511054-3-1-110
55.5 – 58.513557000
58.5 – 61.51156031115
61.5 – 64.525636250
TotalN = 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 – 150150 – 200200 – 250250 -300300 – 350
Number of households:451222

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 expenditureFrequency fiMid value xi  d= xi ­– 225u= (x– 225)50fiui
100 – 1504125-100-2-8
150 – 2005175-50-1-5
200 – 25012225000
250 – 30022755012
300 – 350232510024
 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 SO(in ppm)Frequency
0.00 – 0.044
0.04 – 0.089
0.08 – 0.129
0.12 – 0.162
0.16 – 0.204
0.20 – 0.242

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 SO2Frequency fiClass interval xidi = x– 0.14uifiui
0.00 – 0.0440.02-0.12-3-12
0.04 – 0.0890.06-0.08-2-18
0.08 – 0.1290.1-0.04-1-9
0.12 – 0.1620.14000
0.16 – 0.2040.180.0414
0.20 – 0.2420.220.0824
TotalN = 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 – 66 – 1010 – 1414 – 20 20 – 2828 – 3838 – 40
Number of students:111074431

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 daysNumber of students fiXid = x– 16fidi
0 – 6113-13-143
6 – 10108-8-80
10 – 14712-4-28
14 – 2071600
20 – 28824864
28 – 363331751
30 – 401392323
TotalN = 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 -5555 – 6565 – 7575 – 8585 – 95
Number of cites:3101183

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 xidi= xi ­– 70u= (di –A)10fiui
45 – 55350-20-2– 6
55 – 651060-10-1-10
65 – 751170000
75 – 858801018
85 – 953902026
TotalN = 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 %

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