Question

t has been reported that the average time to download the home page from a government website was 0.9 seconds. Suppose that the download times were normally distributed with a standard deviation of 0.3 seconds. If random samples of 36 download times are selected, what is the probability that the sample mean will be less than 0.84 seconds?

Answer 1

Given,

= 0.9 , = 0.3

Using central limit theorem,

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

So,

P( < 0.84) = P( Z < 0.84 - 0.9 / (0.3 / sqrt(36) ) )

= P( Z < -1.2)

= 1 - P( Z < 1.2)

= 1 - 0.8849 (Probability calculated from Z table)

= 0.1151