Let X be the normally distributed random variable representing the time hiker takes to walk the trail.
Given that mean and standard deviation are 30.1 and 3.6 minutes, respectively,
[tex]\begin{gathered} \mu=30.1 \\ \sigma=3.6 \end{gathered}[/tex]The sample size is 4,
[tex]n=4[/tex]The z-score corresponding to any value of 'x' is given by,
[tex]z=\frac{x-\mu}{\sigma}[/tex]So the probability that the mean time for the sample is less than 28 minutes is calculated as,
[tex]\begin{gathered} P(X<28)=P(z<\frac{28-30.1}{3.6}) \\ P(X<28)=P(z<-0.58) \\ P(X<28)=P(z>0.58) \\ P(X<28)=P(z>0)-P(0From the Standard Normal Distribution Table,[tex]\varnothing(0.58)=0.2190[/tex]Substitute the value,
[tex]\begin{gathered} P(X<28)=0.5-0.2190 \\ P(X<28)=0.281 \\ P(X<28)=28.1\text{ percent} \end{gathered}[/tex]Thus, there is approximately 28.1% probability that the mean time for the sample is less than 28 minutes.
