Question

According to a social media​ blog, time spent on a certain social networking website has a mean of 18 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 4 minutes.

If you select a random sample of 100 sessions, what is the probability that the sample mean is between 17.5 and 18.5 minutes?

Answer 1

Solution :

Given that ,

mean =   = 18

standard deviation = = 4

n = 100

= 18

=  / n= 4 / 100=0.4

P(17.5<     <18.5 ) = P[(17.5-18) /0.4 < ( - ) /   < (18.5-18) / 0.4)]

= P( -1.25< Z <1.25 )

= P(Z < 1.25) - P(Z <-1.25 )

Using z table

=0.8944-0.1056

=0.7888

probability= 0.7888