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Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 1

  • Difficulty Level : Expert
  • Last Updated : 03 Mar, 2021

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 – 15024300 – 35030
150 – 20040350 – 40022
200 – 25033400 – 45016
250 – 30028450 – 5007

Find the average expenditure (in rupees) per household.

Solution:

Let the assumed mean (A) = 275

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Class intervalMid value (xi)d= xi – 275ui = (xi – 275)/50Frequency fifiui
100 – 150125-150-324-72
150 – 200175-100-240-80
200 – 250225-50-133-33
250 – 30027500280
300 – 3503255013030
350 – 40037510022244
400 – 45042515031648
450 – 5004752004728
    N = 200Σ fiui = -35

It’s seen that A = 275 and h = 50

So,

Mean = A + h x (Σfi ui/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 plants:0 – 22 – 44 – 66 – 88 – 1010 – 1212 – 14
Number of house:1215623

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

Solution:

From the given data,



To find the class interval we know that,

Class marks (xi) = (upper class limit + lower class limit)/2

Now, let’s compute xi and fixi by the following

Number of plantsNumber of house (fi)xifixi
0 – 2111
2 – 4236
4 – 6155
6 – 85735
8 – 106954
10 – 1221122
12 – 1431339
TotalN = 20 Σ fiui = 162

Here,

Mean = Σ fiui/N

= 162/ 20

= 8.1

Thus, the mean number of plants in a house is 8.1

We have used the direct method as the values of class mark xi and fi is very small.

Question 3. Consider the following distribution of daily wages of workers of a factory

Daily wages (in ₹)100 – 120120 – 140140 – 160160 – 180180 – 200
Number of workers:12148610

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

Solution:



Let us 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= 50Σ fiui = -12

It’s seen that,

A = 150 and h = 20

So,

Mean = A + h x (Σfi ui/N)

= 150 + 20 x (-12/50)

= 150 – 24/5

= 150 = 4.8

= 145.20

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 – 8080 – 8383 – 86
Number of women:2438742

Solution:



Using the relation (xi) = (upper class limit + lower class limit)/ 2

And, class size of this data = 3

Let the assumed mean (A) = 75.5

So, let’s calculate di, ui, fiui as following:

Number of heart beats per minuteNumber of women (fi)xidi = xi – 75.5ui = (x– 755)/hfiui
65 – 68266.5-9-3-6
68 – 71469.5-6-2-8
71 – 74372.5-3-1-3
74 – 77875.5000
77 – 80778.5317
80 – 83481.5628
83 – 86284.5936
 N = 30   Σ fiui = 4

From table, it’s seen that

N = 30 and h = 3

So, the mean = A + h x (Σfi ui/N)

= 75.5 + 3 x (4/30

= 75.5 + 2/5

= 75.9

Therefore, the 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:

Class interval:0 – 66 – 1212 – 1818 – 2424 – 30
Frequency:681097

Solution:

Let’s consider the assumed mean (A) = 15

Class intervalMid – value xid= x– 15u= (x– 15)/6fifiui
0 – 63-12-26-12
6 – 129-6-18-8
12 – 181500100
18 – 24216199
24 – 3027122714
    N = 40Σ fiui = 3

From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σfi ui/N)

= 15 + 6 x (3/40)

= 15 + 0.45

= 15.45

Question 6. Find the mean of the following frequency distribution:

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

Solution:



Let’s consider the assumed mean (A) = 100

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

From the table it’s seen that,

A = 100 and h = 20

Mean = A + h x (Σfi ui/N)

= 100 + 20 x (61/100)

= 100 + 12.2

= 112.2

Question 7. Find the mean of the following frequency distribution:

Class interval:0 – 88 – 1616 – 2424 – 3232 – 40
Frequency:671089

Solution:

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfi ui/N)

= 20 + 8 x (7/40)

= 20 + 1.4

= 21.4

Question 8. Find the mean of the following frequency distribution:

Class interval:0 – 66 – 1212 – 1818 – 2424 – 30
Frequency:7510126

Solution:

Let’s consider the assumed mean (A) = 15

Class intervalMid – value xid= x– 15u= (x– 15)/6fifiui
0 – 63-12-27-14
6 – 129-6-15-5
12 – 181500100
18 – 2421611212
24 – 3027122612
    N = 40Σ fiui = 5

From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σfi ui/N)

= 15 + 6 x (5/40)



= 15 + 0.75

= 15.75

Question 9. Find the mean of the following frequency distribution:

Class interval:0 – 1010 – 2020 – 3030 – 4040 – 50
Frequency:912151014

Solution:

Let’s consider the assumed mean (A) = 25

Class intervalMid – value xid= x– 25u= (x– 25)/10fifiui
0 – 105-20-29-18
10 – 2015-10-112-12
20 – 302500150
30 – 40351011010
40 – 50452021428
    N = 60Σ fiui = 8

From the table it’s seen that,

A = 25 and h = 10

Mean = A + h x (Σfi ui/N)

= 25 + 10 x (8/60)

= 25 + 4/3

= 79/3 = 26.333

Question 10. Find the mean of the following frequency distribution:

Class interval:0 – 88 – 1616 – 2424 – 3232 – 40
Frequency:591088

Solution:

Let’s consider the assumed mean (A) = 20

Class intervalMid – value xid= x– 20u= (x– 20)/8fifiui
0 – 84-16-25-10
8 – 1612-4-19-9
16 – 242000100
24 – 32284188
32 – 4036162816
    N = 40Σ fiui = 5

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfi ui/N)

= 20 + 8 x (5/40)

= 20 + 1

= 21

Question 11. Find the mean of the following frequency distribution:

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

Solution:

Let’s consider the assumed mean (A) = 20



Class intervalMid – value xid= x– 20u= (x– 20)/8fifiui
0 – 84-16-25-12
8 – 1612-8-16-8
16 – 24200040
24 – 32288139
32 – 4036162214
    N = 20Σ fiui = -9

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfi ui/N)

= 20 + 6 x (-9/20)

= 20 – 72/20

= 20 – 3.6

= 16.4

Question 12. Find the mean of the following frequency distribution:

Class interval:10 – 3030 – 5050 – 7070 – 9090 – 110110 – 130
Frequency:58122032

Solution:

Let’s consider the assumed mean (A) = 60

Class intervalMid – value xid= x–60u= (x– 60)/20fifiui
10 – 3020-40-25-10
30 – 5040-20-18-8
50 – 706000120
70 – 90802012020
90 – 11010040236
110 – 13012060326
    N = 50Σ fiui = 14

From the table it’s seen that,

A = 60 and h = 20

Mean = A + h x (Σfi ui/N)

= 60 + 20 x (14/50)

= 60 + 28/5

= 60 + 5.6

= 65.6

Question 13. Find the mean of the following frequency distribution:

Class interval:25 – 3535 – 4545 – 5555 – 6565 – 75
Frequency:6108124

Solution:

Let’s consider the assumed mean (A) = 50

Class intervalMid – value xid= x– 50u= (x– 50)/10fifiui
25 – 3530-20-26-12
35 – 4540-10-110-10
45 – 55500080
55 – 65601011212
65 – 757020248
    N = 40Σ fiui = -2

From the table it’s seen that,

A = 50 and h = 10

Mean = A + h x (Σfi ui/N)

= 50 + 10 x (-2/40)

= 50 – 0.5

= 49.5




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