Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 1
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 – 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 |
-72 |
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 |
Σ 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 – 2 |
2 – 4 |
4 – 6 |
6 – 8 |
8 – 10 |
10 – 12 |
12 – 14 |
Number of house: |
1 |
2 |
1 |
5 |
6 |
2 |
3 |
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 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 |
|
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 – 120 |
120 – 140 |
140 – 160 |
160 – 180 |
180 – 200 |
Number of workers: |
12 |
14 |
8 |
6 |
10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Let us 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 |
Σ 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 – 68 |
68 – 71 |
71 – 74 |
74 – 77 |
77 – 80 |
80 – 83 |
83 – 86 |
Number of women: |
2 |
4 |
3 |
8 |
7 |
4 |
2 |
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 minute |
Number of women (fi) |
xi |
di = xi – 75.5 |
ui = (xi – 755)/h |
fiui |
65 – 68 |
2 |
66.5 |
-9 |
-3 |
-6 |
68 – 71 |
4 |
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 |
|
|
|
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 – 6 |
6 – 12 |
12 – 18 |
18 – 24 |
24 – 30 |
Frequency: |
6 |
8 |
10 |
9 |
7 |
Solution:
Let’s consider the assumed mean (A) = 15
Class interval |
Mid – value xi |
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 |
12 |
2 |
7 |
14 |
|
|
|
|
N = 40 |
Σ 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 – 70 |
70 – 90 |
90 – 110 |
110 – 130 |
130 – 150 |
150 – 170 |
Frequency: |
18 |
12 |
13 |
27 |
8 |
22 |
Solution:
Let’s consider the assumed mean (A) = 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 |
Σ 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 – 8 |
8 – 16 |
16 – 24 |
24 – 32 |
32 – 40 |
Frequency: |
6 |
7 |
10 |
8 |
9 |
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 – 6 |
6 – 12 |
12 – 18 |
18 – 24 |
24 – 30 |
Frequency: |
7 |
5 |
10 |
12 |
6 |
Solution:
Let’s consider the assumed mean (A) = 15
Class interval |
Mid – value xi |
di = xi – 15 |
ui = (xi – 15)/6 |
fi |
fiui |
0 – 6 |
3 |
-12 |
-2 |
7 |
-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 |
Σ 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 – 10 |
10 – 20 |
20 – 30 |
30 – 40 |
40 – 50 |
Frequency: |
9 |
12 |
15 |
10 |
14 |
Solution:
Let’s consider the assumed mean (A) = 25
Class interval |
Mid – value xi |
di = xi – 25 |
ui = (xi – 25)/10 |
fi |
fiui |
0 – 10 |
5 |
-20 |
-2 |
9 |
-18 |
10 – 20 |
15 |
-10 |
-1 |
12 |
-12 |
20 – 30 |
25 |
0 |
0 |
15 |
0 |
30 – 40 |
35 |
10 |
1 |
10 |
10 |
40 – 50 |
45 |
20 |
2 |
14 |
28 |
|
|
|
|
N = 60 |
Σ 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 – 8 |
8 – 16 |
16 – 24 |
24 – 32 |
32 – 40 |
Frequency: |
5 |
9 |
10 |
8 |
8 |
Solution:
Let’s consider the assumed mean (A) = 20
Class interval |
Mid – value xi |
di = xi – 20 |
ui = (xi – 20)/8 |
fi |
fiui |
0 – 8 |
4 |
-16 |
-2 |
5 |
-10 |
8 – 16 |
12 |
-4 |
-1 |
9 |
-9 |
16 – 24 |
20 |
0 |
0 |
10 |
0 |
24 – 32 |
28 |
4 |
1 |
8 |
8 |
32 – 40 |
36 |
16 |
2 |
8 |
16 |
|
|
|
|
N = 40 |
Σ 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 – 8 |
8 – 16 |
16 – 24 |
24 – 32 |
32 – 40 |
Frequency: |
5 |
6 |
4 |
3 |
2 |
Solution:
Let’s consider the assumed mean (A) = 20
Class interval |
Mid – value xi |
di = xi – 20 |
ui = (xi – 20)/8 |
fi |
fiui |
0 – 8 |
4 |
-16 |
-2 |
5 |
-12 |
8 – 16 |
12 |
-8 |
-1 |
6 |
-8 |
16 – 24 |
20 |
0 |
0 |
4 |
0 |
24 – 32 |
28 |
8 |
1 |
3 |
9 |
32 – 40 |
36 |
16 |
2 |
2 |
14 |
|
|
|
|
N = 20 |
Σ 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 – 30 |
30 – 50 |
50 – 70 |
70 – 90 |
90 – 110 |
110 – 130 |
Frequency: |
5 |
8 |
12 |
20 |
3 |
2 |
Solution:
Let’s consider the assumed mean (A) = 60
Class interval |
Mid – value xi |
di = xi –60 |
ui = (xi – 60)/20 |
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 |
Σ 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 – 35 |
35 – 45 |
45 – 55 |
55 – 65 |
65 – 75 |
Frequency: |
6 |
10 |
8 |
12 |
4 |
Solution:
Let’s consider the assumed mean (A) = 50
Class interval |
Mid – value xi |
di = xi – 50 |
ui = (xi – 50)/10 |
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 |
Σ 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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