c. For a sample of n = 25 scores, what is the probability that the sample mean will be within 10 points of the population mean? In other words, what is p(95 < M < 115)?

Question

For a normal population with u = 105 and o = 20,

a. What is the probability of obtaining a sample mean greater than 115 for a sample of n=

4 scores?

b. What is the probability of obtaining a sample mean greater than 115 for a sample of n

16 scores?

c. For a sample of n = 25 scores, what is the probability that the sample mean will be

within 10 points of the population mean? In other words, what is p(95 < M < 115)?

Answer 1

z-statistic for mean = (M - mu)/(sigma/sqrt(n)), where mu = 105, sigma = 20

(M-105)/(20/sqrt(n) = (M - 105)/n

a) compute P(M > 115)

compute z for M = 115 and n = 4

z-stat = (115-105)/(20/sqrt(4)) = 1

P(M > 115) = P(z > 1) = P(z < -1) = 0.15587

b) compute P(M > 115)

compute z for M = 115 and n = 16

z-stat = (115-105)/(20/sqrt(16)) = 2

P(M > 115) = P(z > 2) = P(z < -2) = 0.0228

c) compute P(95 < M < 115)

compute z for M = 95 and n = 25

z-stat = (95 - 105)/(20/sqrt(25)) = -2.5

compute z for M = 115 and n = 25

z-stat = (115 - 105)/(20/sqrt(25)) = 2.5

P(95 < M < 115) = P(-2.5 < z < 2.5) = P(z < 2.5) - P(z < -2.5) = 0.9938 - 0.0062 = 0.9876