Norminv and Normdist

Your boss says that 27% of customers order wine that costs above a certain amount. What is that amount. Mean 24, SD4.


Notice the word ‘above’. So we want to find the value of the random variable ‘x’....here the price of the wine....that separates the top 27% of the customers from the bottom---and here’s the key---73%. Recall that Excel goes from left to right. We know the percentage, want to find the value that gives that percentage. In this case, norminv is the boy! So go
=norminv(0.73,24,4)=26.45.

Now, a contrasting problem and more complex. In a restaurant, the bills are normally distributed. The mean bill is 28, and the standard deviation is 6. If 12 of the day’s bills are over 43.06 how many customers did the restaurant have that day?
First, find the percentage of the total bills occupied by the area to the right of 43.06. Here we have the value of the random variable and we want to get the area (or probability, whichever way you want to look at it). Recall that Excel adds from left to right....and our area is clearly to the right. It says ‘above’. So go
=1-normdist(43.06,28,6,true) and get 0.006.


This means that 0.6% of the restaurant’s bills were over 43.06.
Now, we know that 0.6% of the total number of bills is 12 in number. Call the total number of bills ‘L’. We are trying to find L.
0.006*L = 12.
Divide both sides by 0.006 to find L
L = 2000.