 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping • Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details

```Chapter 7: Statistics Exercise – 7.2

Question: 1

The number of telephone calls received at an exchange per interval for 250 successive one- minute intervals are given in the following frequency table:

No. of calls(x):
0
1
2
3
4
5
6

No. of intervals (f):
15
24
29
46
54
43
39

Compute the mean number of calls per interval.

Solution:

Let be assumed mean (A) = 3

No. of calls xi
No. of intervals fi
u1 = xi − A = xi = 3
fiui

0
15
- 3
- 45

1
24
- 2
- 48

2
29
- 1
- 29

3
46
0
0

4
54
1
54

5
43
2
86

6
39
3
117

N = 250

Sum = 135

Mean number of cells = 3 + 135/250
= 885/250
= 3.54

Question: 2

Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No of heads per toss (x):
0
1
2
3
4
5

No of tosses (f):
38
144
342
287
164
25

Solution:

Let the assumed mean (A) = 2

No. of heads per toss xi
No of intervals fi
ui = Ai –x = Ai - 2
fiui

0
38
- 2
- 76

1
144
- 1
- 144

2
342
0
0

3
287
1
287

4
164
2
328

5
25
3
75

N = 1000

Sum = 470

Mean number of per toss = 2 + 470/1000
= 2 + 0.47
= 2.47

Question: 3

The following table gives the number of branches and number of plants in the garden of a school.

No of branches (x):
2
3
4
5
6

No of plants (f):
49
43
57
38
13

Calculate the average number of branches per plant.

Solution:

Let the assumed mean (A) = 4

No of branches xi
No of plants fi
ui = xi − A = xi − 4
fiui

2
49
- 2
- 98

3
43
- 1
- 43

4
57
0
0

5
38
1
38

6
13
2
26

N = 200

Sum = – 77

Average number of branches per plant = 4 + (-77/200)
= 4 -77/200
= (800 -77)/200
= 3.615

Question: 4

The following table gives the number of children of 150 families in a village

No of children (x):
0
1
2
3
4
5

No of families (f):
10
21
55
42
15
7

Find the average number of children per family.

Solution:

Let the assumed mean (A) = 2

No of children xi
No of families fi
ui = xi − A = xi − 2
fiui

0
10
- 2
- 20

1
21
- 1
- 21

3
42
1
42

4
15
2
30

5
7
5
35

N = 20

Sum = 52

Average number of children for family = 2 + 52/150
= (300 +52)/150
= 352/150
= 2.35 (approx)

Question: 5

The marks obtained out of 50, by 102 students in a physics test are given in the frequency table below:

Marks (x):
15
20
22
24
25
30
33
38
45

Frequency (f):
5
8
11
20
23
18
13
3
1

Find the average number of marks.

Solution:

Marks xi
Frequency fi
ui = xi  − A = xi − 2
fiui

15
5
- 10
- 50

20
8
- 5
- 40

22
8
- 3
- 24

24
20
- 1
- 20

25
23
0
0

30
18
5
90

33
13
8
104

38
3
12
36

45
3
20
60

N = 122

Sum = 110

Average number of marks = 25 + 110/102
= (2550 + 110)/102
= 2660/102
= 26.08 (Approx)

Question: 6

The number of students absent in a class was recorded every day for 120 days and the information is given in the following

No of students absent (x):
0
1
2
3
4
5
6
7

No of days (f):
1
4
10
50
34
15
4
2

Find the mean number of students absent per day.

Solution:

Let mean assumed mean (A) = 3

No of students absent xi
No of days fi
ui = xi − A = xi − 3
fiui

3
1
- 3
- 3

1
4
- 2
- 8

2
10
- 1
- 10

3
50
0
0

4
34
1
24

5
15
2
30

6
4
3
12

7
2
4
8

N = 120

Sum =63

Mean number of students absent per day = 3 + 63/120
= (360 + 63)/120
= 423/120
= 3.53

Question: 7

In the first proof of reading of a book containing 300 pages the following distribution of misprints was obtained:

No of misprints per page (x):
0
1
2
3
4
5

No of pages (f):
154
96
36
9
5
1

Find the average number of misprints per page.

Solution:

Let the assumed mean (A) = 2

No of misprints per page xi
No of days fi
ui = xi − A = xi − 3
fiui

0
154
- 2
- 308

1
95
- 1
- 95

2
36
0
0

3
9
1
9

4
5
2
1

5
1
3
3

N = 300

Sum = – 381

Average number of misprints per day = 2 + (- 381/300)
= (600-381)/300
= 219/300
= 0.73

Question: 8

Find the mean from the following frequency distribution of marks at a test in statistics:

No of accidents (x):
0
1
2
3
4

No of workers (f):
70
52
34
3
1

Find the average number of accidents per worker.

Solution:

Let the assumed mean (A) = 2

No of accidents
No of workers fi
ui = xi − A = xi − 3
fiui

0
70
- 2
- 140

1
52
- 1
- 52

2
34
0
0

3
3
1
3

4
1
2
2

N = 100

Sum = – 187

Average no of accidents per day workers ⟹ x + (-187/160)
= 133/160
= 0.83

Question: 9

The number of telephone calls received at an exchange per interval for 250 successive one- minute intervals are given in the following frequency table:

Marks (x):
5
10
15
20
25
30
35
40
45
50

No of students (f):
15
50
80
76
72
45
39
9
8
6

Solution:

Let the assumed mean (A) = 25

Marks xi
No of students fi
ui = xi − A = xi − 3
fiui<

5
15
-20
-300

10
50
-15
-750

15
80
-10
-800

20
76
-5
-380

25
72
0
0

30
45
5
225

35
39
10
390

40
9
15
135

45
8
20
160

50
6
25
150

N = 400

Sum = -1170

Mean = 25 + (-1170)/400

= 22.075
```  ### Course Features

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