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  • Complete JEE Main/Advanced Course and Test Series
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Chapter 24: Measures of Central Tendency Exercise – 24.2



Question: 1



Calculate the mean for the following distribution:




	
		
	
	
		

			x:
			5
			6
			7
			8
			9
		

		

			f:
			4
			8
			14
			11
			3
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			5
			4
			20
		

		

			6
			8
			48
		

		

			7
			14
			98
		

		

			8
			11
			88
		

		

			9
			3
			27
		

		

			 
			N = 40
			∑fx = 281
		

	







= 281/40 = 7.025.



 



Question: 2



Find the mean of the following data:




	
		
	
	
		

			x:
			19
			21
			23
			25
			27
			29
			31
		

		

			f:
			13
			15
			16
			18
			16
			15
			13
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			19
			13
			247
		

		

			21
			15
			315
		

		

			23
			16
			368
		

		

			25
			18
			450
		

		

			27
			16
			432
		

		

			29
			15
			435
		

		

			31
			13
			403
		

		

			 
			N = 106
			∑fx = 2650
		

	







= 2650/106 = 25.



 



Question: 3



The mean of the following data is 20.6 .Find the value of p.




	
		
	
	
		

			x:
			10
			15
			p
			25
			35
		

		

			f:
			3
			10
			25
			7
			5
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			10
			3
			30
		

		

			15
			10
			150
		

		

			P
			25
			25p
		

		

			25
			7
			175
		

		

			35
			5
			175
		

		

			 
			N = 50
			∑fx = 25p + 530
		

	




It is given that,



Mean = 20.6






⇒ 25p + 530 = 20.6 × 50



⇒ 25p = 1030 − 530



⇒ 25p = 500



⇒ p = 500/25 = 20



⇒ p = 20



∴ p = 20.



 



Question: 4



If the mean of the following data is 15, find p.




	
		
	
	
		

			f:
			6 
			p
			6
			10
			5
		

		

			x:
			5
			10
			15
			20
			25
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			5
			6
			30
		

		

			10
			P
			10p
		

		

			15
			6
			90
		

		

			20
			10
			200
		

		

			25
			5
			125
		

		

			 
			N = p + 27
			∑fx = 10p + 445
		

	




It is given that,



Mean = 15



⇒∑fx/N = 15



⇒10p + 445p + 27=15



⇒10p + 445 = 15 × (p + 27)



⇒ 10p + 445 = 15p + 405



⇒15p − 10p = 445 − 405



⇒ 5p = 40



⇒ p = 405 = 8



⇒ p = 8



∴ p = 8.



 



Question: 5



Find the value of p for the following distribution whose mean is 16.6.




	
		
	
	
		

			x:
			8
			12
			15
			p
			20
			25
			30
		

		

			f:
			12
			16
			20
			24
			16
			8
			4
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			8
			12
			96
		

		

			12
			16
			192
		

		

			15
			20
			300
		

		

			P
			24
			24p
		

		

			20
			16
			320
		

		

			25
			8
			200
		

		

			30
			4
			120
		

		

			 
			N = 100
			∑fx = 24p + 1228
		

	




It is given that,



Mean = 16.6






⇒ 24p + 1228 = 1660



⇒ 24p = 1660 − 1228



⇒ 24p = 432



⇒ p = 432/24 = 18



⇒ p = 18



∴ p = 18.



 



Question: 6



Find the missing value of p for the following distribution whose mean is 12.58.




	
		
	
	
		

			x:
			5
			8
			10
			12
			p
			20
			25
		

		

			f:
			2
			5
			8
			22
			7
			4
			2
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			5
			2
			10
		

		

			8
			5
			40
		

		

			10
			8
			80
		

		

			12
			22
			264
		

		

			P
			7
			7p
		

		

			20
			4
			80
		

		

			25
			2
			50
		

		

			 
			N = 50
			∑fx = 7p + 524
		

	




It is given that,



Mean = 12.58



⇒ ∑fx/N = 12.58






⇒ 7p + 524 = 629



⇒ 7p = 629 − 524



⇒ 7p = 105



⇒ p = 1057 = 15



⇒ p = 15



∴ p = 18.



 



Question: 7



Find the missing frequency (p) for the following distribution whose mean is 7.68.




	
		
	
	
		

			x
			3
			5
			7
			9
			11
			13
		

		

			f
			6
			8
			15
			p
			8
			4
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			3
			6
			18
		

		

			5
			8
			40
		

		

			7
			15
			105
		

		

			9
			P
			9p
		

		

			11
			8
			88
		

		

			13
			4
			52
		

		

			 
			N = p + 41
			∑fx = 9p + 303
		

	




It is given that,



Mean = 7.68



⇒ ∑fx/N = 7.68



⇒ 9p + 303p + 41 = 7.68



⇒ 9p + 303 = 7.68p + 314.88



⇒ 9p − 7.68p = 314.88 − 303



⇒ 1.32p = 11.88



⇒ p = 11.881.32 = 9



⇒ p = 9



∴ p = 9.



 



Question: 8



Find the value of p, if the mean of the following distribution is 20.




	
		
	
	
		

			x:
			15
			17
			19
			20 + p
			23
		

		

			f:
			2
			3
			4
			5p
			6
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			15
			2
			30
		

		

			17
			3
			51
		

		

			19
			4
			76
		

		

			20+p
			5p
			100p + 5p2
		

		

			23
			6
			138
		

		

			 
			N = 5p + 15
			Fx = 5p2 + 100p + 295
		

	




It is given that,



Mean = 20






⇒ 5p2 + 100p + 295 = 20(5p + 15)



⇒ 5p2 + 100p + 295 = 100p + 300



⇒ 5p2 = 300 − 295



⇒ 5p2 = 5



⇒ p2 = 1



⇒ p = ±1



Frequency can’t be negative.



Hence, value of p is 1.



 



Question: 9



Find the mean of the following distribution:




	
		
	
	
		

			x:
			10
			12
			20
			25
			35
		

		

			f:
			3
			10
			15
			7
			5
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			10
			3
			30
		

		

			12
			10
			120
		

		

			20
			15
			300
		

		

			25
			7
			175
		

		

			35
			5
			175
		

		

			 
			N = 40
			∑fx = 800
		

	







= 800/40 = 20.



 



Question: 10



Candidates of four schools appear in a mathematics test. The data were as follows:




	
		
	
	
		

			Schools
			No. of Candidates
			Average Score
		

		

			I
			60
			75
		

		

			II
			48
			80
		

		

			III
			Not Available
			55
		

		

			IV
			40
			50
		

	




If the average score of the candidates of all four schools is 66, Find the number of candidates that appeared from school III. 



Solution:




	
		
	
	
		

			Schools
			No. of Candidates
			Average Score
		

		

			I
			60
			75
		

		

			II
			48
			80
		

		

			III
			x
			55
		

		

			IV
			40
			50
		

	




Given the average score of all schools = 66






⇒ 10340 + 55x = 66x + 9768



⇒ 10340 − 9768 = 66x − 55x



⇒ 11x = 572



⇒ x = 572/11 = 52



∴ No. of candidates appeared from school III = 52.



 



Question: 11



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
			No. of tosses
		

		

			0
			38
		

		

			1
			144
		

		

			2
			342
		

		

			3
			287
		

		

			4
			164
		

		

			5
			25
		

		

			Total
			1000
		

	




Solution:




	
		
	
	
		

			No. of heads per toss (x)
			No. of tosses (f)
			fx
		

		

			0
			38
			0
		

		

			1
			144
			144
		

		

			2
			342
			684
		

		

			3
			287
			861
		

		

			4
			164
			656
		

		

			5
			25
			125
		

		

			 
			N = 1000
			∑fx = 2470
		

	




∴ Mean number of heads per toss = ∑fx/N



= 2470/1000



= 2.47



 



Question: 12



Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.




	
		
	
	
		

			x:
			10
			30
			50
			70
			90
		

		

			f:
			17
			f1
			32
			f2
			19
		

	




Solution:




	
		
	
	
		

			x
			f
			fx
		

		

			10
			17
			170
		

		

			30
			f1
			30f1
		

		

			50
			32
			1600
		

		

			70
			f2
			70f2
		

		

			90
			19
			1710
		

		

			 
			N = 120
			∑fx = 3480 + 30f1 + 70f2
		

	




It is given that



Mean = 50



⇒ ∑fx/N = 50






⇒ 3480 + 30f1 + 70f2 = 50 × 120



⇒30f1 + 70f2 = 6000 − 3480



⇒10(3f1 + 7f2) = 10(252)



⇒ 3f1 + 7f2 = 252 ⋅⋅⋅⋅ (1)  [∵ Divide by 10]



And N = 20



⇒ 17 + f1 + 32 + f2 + 19 = 120



⇒ 68 + f1 + f2 = 120



⇒ f1 + f2 = 120 − 68



⇒ f1 + f2 = 52



Multiply with 3 on both sides



⇒ 3f1 + 3f2 = 156 ⋅⋅⋅ (2)



Subtracting equation (2) from equation (1)



⇒ 3f1 + 7f2 − 3f1 − 3f2 = 252 − 156



⇒ 4f2 = 96



⇒ f2 = 96/4 = 24



Put the value of f2 in equation (1)



⇒ 3f1 + 7 × 24 = 252



⇒ 3f1 = 252 − 168



⇒ f1 = 84/3 = 28



⇒ f1 = 28


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