Question

Suppose that the heights of male students at a university have a normal distribution with mean = 65 inches and standard deviation = 2.0 inches. A randomly sample of 10 students are selected to make up an intramural basket ball team.

i) What is the mean (mathematical expectation) of xbar?

ii) What is the standard deviation of x bar?

iii) What is the probability that the average height (x bar) of the team will exceed 69 inches?

iv) What is the probability that the average height (x bar) of the team will be between 62 and 70 inches?

Answer 1

Given: = 65 inches, = 2 inches, n = 10

(i) The mean of = = 65

(ii) The standard deviation of = = =

(iii) To Find probability, we need to find the Z scores.

Z = (X - )/

P(X > 69) = 1 - P(X < 69)

For P(X < 69), Z = (69 - 65)/0.6325 = 6.32

The Probability of X < 69 = 1   (From the normal distribution tables)

Therefore the probability of X > 69 = 1 - 1 = 0

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To Find P( 62 < X < 70) = P(X < 70) - P(X < 62)

For P(X < 70), Z = (70 - 65)/0.6325 = 7.91

The Probability of X < 70 = 1   (From the normal distribution tables)

For P(X < 62), Z = (62 - 65)/0.6325 = -4.74

The Probability of X < 62 = 0 (From the normal distribution tables)

Therefore the required probability = 1 - 0 = 1

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