• 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

• Complete AIPMT/AIIMS Course and Test Series
• OFFERED PRICE: Rs. 15,900


• View Details Xpress Buy

Course Features

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