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 | âˆ‘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 | âˆ‘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 | âˆ‘fx=25p+530 |
It is given that ,
Mean = 20.6
â‡’ 25p+530 = 20.6Ã—50
â‡’ 25p = 1030âˆ’530
â‡’ 25p = 500
â‡’ p = 50025 = 20
â‡’ p = 20
âˆ´ 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 | âˆ‘fx = 10p+445 |
It is given that ,
Mean = 15
â‡’ 10p+445 = 15Ã—(p+27)
â‡’10p+445 = 15p+405
â‡’15pâˆ’10p = 445âˆ’405
â‡’ 5p = 40
â‡’ p = 40/5 =8
â‡’ p = 8
âˆ´ 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 | âˆ‘fx = 24p+1228 |
It is given that ,
Mean = 16.6
â‡’ 24p+1228 = 1660
â‡’ 24p = 1660âˆ’1228
â‡’ 24p = 432
â‡’ p = 432/24 = 18
â‡’ p = 18
âˆ´ 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 | âˆ‘fx=7p+524 |
It is given that ,
Mean = 12.58
â‡’ 7p+524 = 629
â‡’ 7p = 629âˆ’524
â‡’ 7p = 105
â‡’ p = 1057 = 15
â‡’ p = 15
âˆ´ 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 | âˆ‘fx=9p+303 |
It is given that ,
Mean = 7.68
â‡’ 9p+303 = 7.68p+314.88
â‡’ 9pâˆ’7.68p = 314.88âˆ’303
â‡’ 1.32p = 11.88
â‡’ p = 11.881.32 = 9
â‡’ p = 9
âˆ´ 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
â‡’ 5p^{2}+100p+295 = 20(5p+15)
â‡’ 5p^{2}+100p+295 = 100p+300
â‡’ 5p^{2} =300âˆ’295
â‡’ 5p^{2} = 5
â‡’ p^{2} = 1
â‡’ p = Â±1
Frequency canâ€™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 | âˆ‘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
â‡’ 10340+55x = 66x+9768
â‡’ 10340âˆ’9768 = 66xâˆ’55x
â‡’ 11x = 572
â‡’ x = 572/11 = 52
âˆ´ 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 | âˆ‘fx = 2470 |
âˆ´ 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 | âˆ‘fx = 3480+30f_{1}+70f_{2} |
It is given that
Mean = 50
â‡’ 3480+30f_{1}+70f_{2 }= 50Ã—120
â‡’ 30f_{1}+70f_{2 }= 6000âˆ’3480
â‡’10(3f_{1}+7f_{2}) = 10(252)
â‡’3f_{1}+7f_{2 }= 252â‹…â‹…â‹…â‹…â‹…(1) [âˆµDivideby10]
And N = 20
â‡’17+f_{1}+32+f_{2}+19 = 120
â‡’ 68+f_{1}+f_{2 }= 120
â‡’ f_{1}+f_{2 }= 120âˆ’68
â‡’ f_{1}+f_{2 }= 52
Multiply with 3 on both sides
â‡’3f_{1}+3f_{2 }= 156â‹…â‹…â‹…â‹…â‹…â‹…(2)
Subtracting equation (2) from equation (1)
â‡’3f_{1}+7f_{2}âˆ’3f_{1}âˆ’3f_{2 }= 252âˆ’156
â‡’ 4f_{2}=96
â‡’ f_{2 }= 96/4 = 24
Put the value of f_{2} in equation (1)
â‡’ 3f_{1}+7Ã—24 = 252
â‡’ 3f_{1 }= 252âˆ’168
â‡’ f_{1 }= 84/3 = 28
â‡’ f_{1 }= 28