Question

According to a social media blog, time spent on a certain social networking website has a mean of 20 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 7 minutes. Complete parts (a) through (d) below. a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 19.5 and 20.5 minutes? (Round to three decimal places as needed.)

Answer 1

Note that sample mean X follows a Normal distribution with Mean : 1-20 Standard deviation - o/Vn-1.4 Note that z: (x-μ)/(o/vn) -(x-20)/1.4 --(Eq.1) follows standard Normal. When 19.5, from (Eq.1),z -0.36 When- 20.5, from (Eq.1), z 0.36 P(19.5 20.5)-P(-0.36Z0.36)- P(Z0.36) - P(Z -0.36)-0.6406-0.3594 0.2812 (ans)