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 a random sample of 25 ​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 = 25

= 18

=  / n= 4 / 25=0.8

P(17.5<     <18.5 ) = P[(17.5-18) /0.8< ( - ) /   < (18.5-18) /0.8 )]

= P( -0.625< Z < 0.625)

= P(Z < 0.625) - P(Z <-0.625 )

Using z table

=0.7340-0.2660

=0.4680

probability= 0.4680