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USE CODE: SELF10 

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

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