Step 1 (a)

Consider a random variable X which represent time taken by mechanic to replace the clutch

The probability that the car will be ready in two hours can be calculated as:

\(\displaystyle{P}{\left({X}{<}{120}\right)}={P}{\left({Z}{<}{\frac{{{120}-{90}}}{{{15}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{<}{2}\right)}\)

\(\displaystyle={0.97725}\) (from standard normal table)

Thus, the required probability is 0.97725.

Step 2 (b)

The probability that the car will be not be ready after 80 minutes can be calculated as:

\(\displaystyle{P}{\left({X}{>}{80}\right)}={1}-{P}{\left({X}\le{80}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}\le{\frac{{{80}-{90}}}{{{15}}}}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}\le-{0.66667}\right)}\)

\(\displaystyle={1}-{0.252482}\) (from standard normal table)

\(\displaystyle={0.747518}\)

Thus, the required probability is 0.7475.

Step 3

The probability that the car will be ready in 85 to 95 minutes can be calculated as:

\(\displaystyle{P}{\left({85}{<}{X}{<}{95}\right)}={P}{\left({\frac{{{85}-{90}}}{{{15}}}}{<}{Z}{<}{\frac{{{95}-{90}}}{{{15}}}}\right)}\)

\(\displaystyle={P}{\left(-{0.333}{<}{Z}{<}{0.333}\right)}\)

\(\displaystyle={P}{\left({Z}{<}{0.333}\right)}-{P}{\left({Z}\le-{0.3333}\right)}\)

\(\displaystyle={0.630433}-{0.369567}\) (from standard normal table)

\(\displaystyle={0.260866}\)

So the probability is 0.2609.