weekly cleaning budget

Question

The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean 𝜇 = $623 and standard deviation 𝜎 = $40.

How much should be budgeted for weekly repairs, cleaning, and maintenance so that the likelihood of the budgeted amount being exceeded in a given week is only 11% ?

$  Round answer to nearest whole dollar.

Answer 1

We want to find the amount budgeted so the probability of it being exceeded is 11%. So the probability that it is not exceeded is 89%.

Since we know the probability and we want to know the number of dollars, we use inverse-normal.

One way to use it is to find the z-score such that 89% is below and 11% is above:

Inverse-norm takes the area to the left as the input and returns the z-score:

invnorm(0.89) = 1.23

So our z-score is 1.23. And z = (x - u) / sigma

we know u=623 and sigma=40 and want x:

1.23 = (x-623)/40 so x=672

You could also use invnorm(0.89,623,40) and let the calculator deal with the z-score