Question

To estimate the mean height μ of male students on your campus, you will measure an SRS of students. Heights of people of the same sex and similar ages are close to Normal. You know from government data that the standard deviation of the heights of young men is about 2.8 inches. Suppose that (unknown to you) the mean height of all male students is 70 inches. (a) If you choose one student at random, what is the probability that he is between 67 and 72 inches tall? (b) You measure 25 students. What is the standard deviation of the sampling distribution of their average height x⎯⎯⎯? (c) What is the probability that the mean height of your sample is between 67 and 72 inches?

Answer 1

Given,

= 70, = 2.8

We convert this to standard normal as

P( X < x) = P (Z < x - / )

a)

p( 67 < X < 72) = P( X < 72) - P( X < 67)

= P( Z < 72 - 70 / 2.8) - P(Z < 67 - 70 / 2.8)

= P( Z < 0.7143) - P( Z < -1.0714)

= 0.7625 - 0.1420

= 0.6205

b)

Standard deviation of sample mean = / sqrt(n) = 2.8 / sqrt (25) = 0.56

c)

Using central limit theorem,

P( < x) = P( Z < x - / / sqrt(n) )

Therefore,

p( 67 < < 72) = P( < 72) - P( < 67)

= P( Z < 72 - 70 / 2.8 / sqrt(25) ) - P(Z < 67 - 70 / 2.8 / sqrt(25) )

= P( Z < 3.5714) - P( Z < -5.3571)

= 0.9998 - 0

= 0.9998