Step 1

Solution:

Let X be the delivery times that follows normal distribution with mean \(\displaystyle\mu={48.2}\) minutes and standard deviation \(\displaystyle\sigma={13.4}\) minutes.

Step 2

Probability that the mean time of 5 deliveries will exceed one hour (60 minutes):

The probability that the mean time of 5 deliveries will exceed 60 minutes can be calculated as given below:

\(\displaystyle{P}{\left(\overline{{{x}}}{>}{60}\right)}={P}{\left({\frac{{\overline{{{x}}}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{>}{\frac{{{60}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}\right)}\)

\(\displaystyle={P}{\left({z}{>}{\frac{{{60}-{48.2}}}{{{\frac{{{13.4}}}{{\sqrt{{{5}}}}}}}}}\right)}\)

\(\displaystyle={P}{\left({z}{>}{1.97}\right)}\)

\(\displaystyle={1}-{P}{\left({z}{<}{1.97}\right)}\)

\(\displaystyle={1}-{0.9756}\) (Using standard normal table, \(\displaystyle{P}{\left({z}{<}{1.97}\right)}={0.9756}{)}\)

\(\displaystyle={0.0244}\)

Thus, the probability that the mean time of 5 deliveries will exceed 60 minutes is 0.0244.