Question

Time spent using​ e-mail per session is normally​ distributed, with mu equals 7 minutes and sigma equals 2 minutes. Assume that the time spent per session is normally distributed. Complete parts​ (a) through​ (d). If you select a random sample of 200 ​sessions, what is the probability that the sample mean is between 6.8 and 7.2 ​minutes?

Answer 1

Solution :

Given that,

mean = = 7

standard deviation = = 2

n=200

= 7

=  / n = 2 / 200=0.14

= P(6.8<    < 7.2) = P[(6.8-7) / 0.14< ( - ) / < (7.2-7) /0.14 )]

= P( -1.43< Z <1.43 )

= P(Z <1.43 ) - P(Z <-1.43 )

Using z table,  

= 0.9236-0.0764

= 0.8472