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:
[tex]P(13.6\leq \bar{x} \leq 14.4)[/tex]
First we have to find the z scores as:
[tex]P \left(\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \leq z \leq \frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}} \right)[/tex]
[tex]P \left(\frac{13.6-14}{\frac{4}{\sqrt{25}}} \leq z \leq \frac{14.4-14}{\frac{4}{\sqrt{25}}}} \right)[/tex]
[tex]P(-0.5\leq z \leq 0.5)[/tex]
Using the standard normal table, we have:
[tex]P(-0.5\leq z \leq 0.5)=P(z\leq 0.5)-P(z\leq -0.5)[/tex]
[tex]=0.6915-0.3085[/tex]
[tex]=0.3830[/tex]
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
