part a

Obtain the probability that one of these cars will have an mpg of more than 37.5.

The probability that one of these cars will have an mpg of more than 37.5 is obtained below:

Let X denotes the mileage of cars follows normal distribution with mean 36.8 mileage per gallon and standard deviation is 1.3 mile age gallon. That is, \(\displaystyle\mu={36.8},\sigma={1.3}\).

The required probability is,

\(\displaystyle{P}{\left({X}{>}{37.5}\right)}={1}-{P}{\left({X}\le{37.5}\right)}\)

\(\displaystyle={1}-{P}{\left({\frac{{{X}-\mu}}{{\sigma}}}\le{\frac{{{37.5}-{36.8}}}{{{1.3}}}}\right)}\)

\(\displaystyle={1}-{P}{\left({z}\le{\frac{{{0.7}}}{{{1.3}}}}\right)}\)

\(\displaystyle={1}-{P}{\left({z}\le{0.54}\right)}\)

From the "standard normal table", the area to the left of \(\displaystyle{z}={0.54}\ {i}{s}\ {0.7054}\).

\(\displaystyle{P}{\left({X}{>}{37.5}\right)}={1}-{P}{\left({z}\le{0.54}\right)}\)

\(\displaystyle={1}-{0.7054}\)

\(\displaystyle={0.2946}\)

Thus, the probability that one of these cars will have an mpg of more than 37.5 is 0.2946.

part b

Obtain the probability that one of these cars will have an mpg of less than 35.

The probability that one of these cars will have an mpg of less than 35 is obtained below:

The required probability is,

\(\displaystyle{P}{\left({X}{<}{35}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}\le{\frac{{{35}-{36.8}}}{{{1.3}}}}\right)}\)

\(\displaystyle={P}{\left({z}\le{\frac{{-{1.8}}}{{{1.3}}}}\right.}\)

\(\displaystyle={P}{\left({z}\le-{1.38}\right)}\)

From the "standard normal table", the area to the left of \(\displaystyle{z}=-{1.38}\ {i}{s}\ {0.0838}\).

\(\displaystyle{P}{\left({X}{<}{35}\right)}={P}{\left({z}\le-{1.38}\right)}\)

\(\displaystyle={0.0838}\)

Thus, the probability that one of these cars will have an mpg of less than 35 is 0.0838.