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Q 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 6 7 8 9  4 8 14 11 3  20 48 98 88 27 
N=40  ∑fx = 281 
Q 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 21 23 25 27 29 31  13 15 16 18 16 15 13  247 315 368 450 432 435 403 
N=106  ∑fx=2650 
Q 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 15 P 25 35  3 10 25 7 5  30 150 25p 175 175 
N = 50  ∑fx=25p+530 
It is given that ,
Mean = 20.6
⇒ 25p+530 = 20.6×50
⇒ 25p = 1030−530
⇒ 25p = 500
⇒ p = 50025 = 20
⇒ p = 20
∴ p = 20.
Q 4 . If the mean of the following data is 15 , find p.
x :  5  10  15  20  25 
f :  6  p  6  10  5 
SOLUTION :
x  f  fx 
5 10 15 20 25  6 P 6 10 5  30 10p 90 200 125 
N = p+27  ∑fx = 10p+445 
It is given that ,
Mean = 15
⇒ 10p+445 = 15×(p+27)
⇒10p+445 = 15p+405
⇒15p−10p = 445−405
⇒ 5p = 40
⇒ p = 40/5 =8
⇒ p = 8
∴ p = 8.
Q 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 15 P 20 25 30  12 16 20 24 16 8 4  96 192 300 24p 320 200 120 
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.
Q 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 8 10 12 P 20 25  2 5 8 22 7 4 2  10 40 80 264 7p 80 50 
N = 50  ∑fx=7p+524 
It is given that ,
Mean = 12.58
⇒ 7p+524 = 629
⇒ 7p = 629−524
⇒ 7p = 105
⇒ p = 1057 = 15
⇒ p = 15
∴ p = 18.
Q7 . 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 5 7 9 11 13  6 8 15 P 8 4  18 40 105 9p 88 52 
N=p+41  ∑fx=9p+303 
It is given that ,
Mean = 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.
Q 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 17 19 20+p 23  2 3 4 5p 6  30 51 76 100p+ 5p2 138 
N=5p+15  fx= 5p^{2}+100p+295 
It is given that ,
Mean = 20
⇒ 5p^{2}+100p+295 = 20(5p+15)
⇒ 5p^{2}+100p+295 = 100p+300
⇒ 5p^{2} =300−295
⇒ 5p^{2} = 5
⇒ p^{2} = 1
⇒ p = ±1
Frequency can’t be negative.
Hence, value of p is 1.
Q 9 . Find the mean of the following distribution :
x :  10  12  20  25  30 
f :  3  10  15  7  5 
SOLUTION :
x  f  fx 
10 12 20 25 35  3 10 15 7 5  30 120 300 175 175 
N = 40  ∑fx = 800 
Q 10. Candidates of four schools appear in a mathematics test. The data were as follows :
Schools  No. Of Candidates  Average Score 
I II III IV  60 48 Not Available 40  75 80 55 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 II III IV  60 48 x 40  75 80 55 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.
Q.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 1 2 3 4 5  38 144 342 287 164 25 
Total  1000 
SOLUTION :
No . of heads per toss(x)  No.of tosses(f)  fx 
0 1 2 3 4 5  38 144 342 287 164 25  0 144 684 861 656 125 
N = 1000  ∑fx = 2470 
∴ Mean number of heads per toss
= 2470/1000
= 2.47
Q 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  f_{1}  32  f_{2}  19 
Total = 120
SOLUTION :
x  f  fx 
10
30 50 70 90  17
f1 32 f2 19  170
30f1 1600 70f2 1710 
N=120  ∑fx = 3480+30f_{1}+70f_{2} 
It is given that
Mean = 50
⇒ 3480+30f_{1}+70f_{2 }= 50×120
⇒ 30f_{1}+70f_{2 }= 6000−3480
⇒10(3f_{1}+7f_{2}) = 10(252)
⇒3f_{1}+7f_{2 }= 252⋅⋅⋅⋅⋅(1) [∵Divideby10]
And N = 20
⇒17+f_{1}+32+f_{2}+19 = 120
⇒ 68+f_{1}+f_{2 }= 120
⇒ f_{1}+f_{2 }= 120−68
⇒ f_{1}+f_{2 }= 52
Multiply with 3 on both sides
⇒3f_{1}+3f_{2 }= 156⋅⋅⋅⋅⋅⋅(2)
Subtracting equation (2) from equation (1)
⇒3f_{1}+7f_{2}−3f_{1}−3f_{2 }= 252−156
⇒ 4f_{2}=96
⇒ f_{2 }= 96/4 = 24
Put the value of f_{2} in equation (1)
⇒ 3f_{1}+7×24 = 252
⇒ 3f_{1 }= 252−168
⇒ f_{1 }= 84/3 = 28
⇒ f_{1 }= 28
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