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.

a. If you select random sample of 25 sessions, what is the probability that the sample mean is between 17 and 18 minutes?

Answer 1

Solution :

Given that ,

mean =   = 18

standard deviation = = 4

n = 25

= 18

=  / n= 4/ 25=0.8

P(17<     < 18) = P[(17-18) / 0.8< ( - ) /   < (18-18) /0.8 )]

= P( -1.25< Z < 0)

= P(Z < 0) - P(Z < -1.25)

Using z table

=0.5-0.1056

=0.3944

probability= 0.3944