Question

According to a social media​ blog, time spent on a certain social networking website has a mean of 16 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 5 minutes.

If you select a random sample of 36 ​sessions, what is the probability that the sample mean is between 15.5 and 16.5 minutes?

(Round your answer to three decimal places)

Answer 1

Using central limit theorem,

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

So,

P(15.5 < < 16.5) = P( < 16.5) - P( < 15.5)

= P( Z < 16.5 - 16 / 5 / sqrt(36) ) - P( Z < 15.5 - 16 / 5 / sqrt(36) )

= P( Z < 0.6) - P( Z < -0.6)

= P( Z < 0.6 ) - ( 1 - P( Z < 0.6)

= 0.7257 - ( 1 - 0.7257)

= 0.452