# The amount of time required for a mechanic to replace a clutch is norm

The amount of time required for a mechanic to replace a clutch is normally distributed with mean 90 minutes and standard deviation of 15 minutes. A car is taken to the mechanic to have its clutch replaced.
a) Find the probability that the car will be ready in two hours.
b) Find the probability that the car will not be ready after 80 minutes.
c) Find the probability that the car is ready in between 85 to 95 minutes.

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Befory

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.