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Calculate the mean for the following distribution:
Mean = 281/40 =7.025
Find the mean of the following data:
If the mean of the following data is 20.6. Find the value of p.
Given Mean = 20.6
20.6 = (530 + 25p)/50
20.6 25p = 20.6 P
P = 20
If the mean of the following data is 15, find p
Given Mean =15
15P + 405 = 10P + 445
15P - 10P = 445-405
5P = 40
P = 5
Find the value of p for the following distribution whose mean is 16.6
Given Mean = 16.6
16.6 = (24p +1228)/100
1660 = 24p + 1228
24p = 1660 – 1228
P = 432/24
P = 18
Find the missing value of p for the following distribution whose mean is 12.58
Given mean = 12.58 Mean = Sum/N 12.58 = (524 +7p)/50 629 = 524 + 7p 629 - 524 = 7p 105 = 7P P = 105/7 P = 15
Find the missing frequency (p) for the following distribution whose mean is 7.68.
Given Mean = 7.68
7.68 (P + 41) = 9P + 303 7.68 P + 314.88 = 9P + 303 9P – 7.68 P = 314.88 – 303 1.32 P = 11.88 P = 11.88/1.32 P = 9
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students.
Mean age = sum/ N = 698/ 40 = 17.45 years
Candidates of four schools appear in a mathematics test. The data were as follows:
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Let the number candidates from school III = P
Given Average score or all schools = 66
10340 + 55p = 66 p + 9768 10340 - 9768 = 66p –55 p P = 572/11 P = 52
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.
Mean number of heads per toss = 2470/1000 = 2.47 Mean = 2.47
The arithmetic mean of the following data is 25. Find the value of k.
Given mean = 25 Sum/ N = 25
25K + 350 = 15K + 390 25 K – 15K = 390 - 350 10 k = 40 K = 40/10 K = 4
If the mean of the following data is 18.75. Find the value of p.
Given mean = 18.75
18.75p + 468.75 = 20p + 460 468.75 – 460 = 20p – 18.75p 8.75 = 1.25P P = 8.75/1.25 P = 7
Find the value of p. If the mean of the following distribution is 20.
Given Mean = 20
100p + 300 = 295 + 100p + 5p2 300 – 295 = 5p2 + 100p 5 = 5p2 + 100P 5p2 + 100P – 5 =0 5p2 + 20p –5P 5(p2 – 1) = 0 P2 – 1 = 0 P (+1, -1) If p + 1 = 0, p = -1, Or p – 1 = 0, P = 1
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50. If ∑f = 120
Given mean = Sum/N = 50
3400 + 50 f1 + 50 f2 = 30 f1 + 70 f2 + 3480 50 f1 – 30 f1 + 50 f2 – 70 f2 = 3480 – 3400 20 f1 – 20 f2 = 80 f1 – f2 = 4 ….. (i) f1 + f2 + 68 = 120 f1 + f2 = 52 …. (ii) Add both equation f1 = 28 f2 = 24
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