a) P(M > 110) = P(Z > (110 - 100)/(20/sqrt(4))) = P(Z > 1)
b) P(M > 110) = P(Z > (110 - 100)/(20/sqrt(16))) = P(Z > 2)
c) P(95 < M < 105) = 1 - 2 * P (Z > (105 - 100)/(20/sqrt(25)))
= 1 - 2 * P(Z > 1.25)
Kirsten C.
asked 03/30/20For a normal population with μ = 100 and σ = 20, find the following.
(a) What is the probability of obtaining a sample mean greater than 110 for a sample of n = 4 scores?
(b) What is the probability of obtaining a sample mean greater than 110 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 5 points of the population mean? In other words, what is p(95 < M < 105)?
a) P(M > 110) = P(Z > (110 - 100)/(20/sqrt(4))) = P(Z > 1)
b) P(M > 110) = P(Z > (110 - 100)/(20/sqrt(16))) = P(Z > 2)
c) P(95 < M < 105) = 1 - 2 * P (Z > (105 - 100)/(20/sqrt(25)))
= 1 - 2 * P(Z > 1.25)
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