Question

According to a mashable post, time spent on tumblr, a microblogging platform and social networking website, has a mean of 14 minutes per visit. (source: on.Mash.To/1757wfe.) as- sume that time spent on tumblr per visit is normally distributed and that the standard deviation is 4 minutes. If you select a random sample of 25 visits,

a. What is the probability that the sample mean is between 13.6 and 14.4 minutes

Answer 1

Solution: We are given that the time spent on tumblr, a microblogging platform and social networking website is normally distributed with mean = 14 minutes and standard deviation = 4 minutes

We have to find, If we select a random sample of 25 visits, the probability that the sample mean is between 13.6 and 14.4 minutes. In other words, we have to find:

`P(13.6\leq \bar{x} \leq 14.4)`

First we have to find the z scores as:

`P \left(\frac{\bar{x}-\mu}{(\sigma)/(√(n))} \leq z \leq \frac{\bar{x}-\mu}{(\sigma)/(√(n))}} \right)`

`P \left((13.6-14)/((4)/(√(25))) \leq z \leq (14.4-14)/((4)/(√(25)))} \right)`

`P(-0.5\leq z \leq 0.5)`

Using the standard normal table, we have:

`P(-0.5\leq z \leq 0.5)=P(z\leq 0.5)-P(z\leq -0.5)`

`=0.6915-0.3085`

`=0.3830`

Therefore, if a random sample of 25 visits is selected, the probability that the sample mean is between 13.6 and 14.4 minutes is 0.3830