The amount of time required for a mechanic to replace a clutch is normally distributed...

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:
${\displaystyle P\left(X>80\right)=1-P(X\le 80)}$
${\displaystyle =1-P(Z\le \frac{80-90}{15})}$
${\displaystyle =1-P(Z\le -0.66667)}$
${\displaystyle =1-0.252482}$ (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:
${\displaystyle P\left(85<X<95\right)=P\left(\frac{85-90}{15}<Z<\frac{95-90}{15}\right)}$
${\displaystyle =P(-0.333<Z<0.333)}$
${\displaystyle =P\left(Z<0.333\right)-P(Z\le -0.3333)}$
${\displaystyle =0.630433-0.369567}$ (from standard normal table)
${\displaystyle =0.260866}$
So the probability is 0.2609.