Question

Assume the time to change the oil and filter in a car has a mean of 25 minutes and a standard deviation of 4 minutes. What is the probability that the total time spent to change the oil and filter in 40 cars is over 1030 minutes?

Answer 1

We want the probability that
$$
\sum_{i=1}^{40}t_i\ge1030\text{ minutes}
$$
Where \(t_i\) is the time to change the oil of car \(i\)
So
$$
p=P\left (\sum_{i=1}^{40}t_i\ge1030\right)=P\left (\cfrac{\sum_{i=1}^{40}t_i}{n}\ge\cfrac{1030}{n}\right)=P\left (\cfrac{\sum_{i=1}^{40}t_i}{n}-\mu\ge\cfrac{1030}{n}-\mu\right)\\
=P\left (\cfrac{\cfrac{\sum_{i=1}^{40}t_i}{n}-\mu}{\sigma }\ge\cfrac{\cfrac{1030}{n}-\mu}{\sigma}\right)=P\left (X\ge \cfrac{\cfrac{1030}{40}-25}{4}\right)
$$
Where \(X\) is normally distributed with mean \(\mu=0\) and standard deviation \(\sigma=1\) after this normalization.
See https://en.wikipedia.org/wiki/Normal_distribution
Then
$$
p=P\left (X\ge \cfrac{3}{16}\right )=0.4256
$$
See \(\href{http:///www.wolframalpha.com/input/?i=+normal+distribution+calculator&f1=-3%2F16&f=NormalProbabilities.z_-3%2F16&a=*FVarOpt.1-_***NormalProbabilities.z--.***NormalProbabilities.pr--.**NormalProbabilities.l-.*NormalProbabilities.r---.*--&a=*FVarOpt.2-_**-.***NormalProbabilities.mu--.**NormalProbabilities.sigma---.**NormalProbabilities.z---}{Normal~Distribution~Calculator}\)