How many of these components should the garage hold, in order that there is a probability of 99% that all requests are satisfied within the day?

Question

I have a probability and statistics question. The question is from a Master's course. The scenario is the following:

A repair garage can order a particular component overnight for delivery at the start of the working day. The demand for this component is normally distributed with a mean of 50 components/day and a standard deviation of 15 components/day.

How many of these components should the garage hold at the beginning of the day, in order that there is a probability of 99% that all requests are satisfied within the day?

My thoughts so far:

99% Confidence Interval = x̅ ± 2.58 Sx̅ x̅ = 50 Sx̅ = 15

1. 50 + (2.58 x 15) = 88.7
2. 50 - (2.58 x 15) = 11.3

I thought the answer should be 88, but according to the course the correct answer is 85. I am not sure how to get that number.

Many thanks.

Answer 1

this is a one tailed problem. You have to calculate

$$P(X\leq x)\geq 0.99$$

thus your Standard gaussian quantile is 2.326 and not 2.58

As a consequence follows that the requested minimum quantity is

$$50+15\times2.3263\approx 84.895=85$$

Answer 2

Letting $X$ denote the demand for the component, you are then looking for the smallest possible integer $n$ such that $$ P(X\leq n)\geq 0.99. $$ What your confidence interval tells you is that $P(11.3\leq X\leq88.7)=0.99$. So, taking $n=89$ certainly gives you the inequality $P(X\leq 89)\geq 0.99$, however this is likely not the smallest possible value of $n$.

To find the smallest possible $n$ which satisfies the inequality above you could start by solving the equation $P(X\leq x)=0.99$. Here you will need to use the parameters $\mu=50$ and $\sigma=15$ in the normal distribution.