Example:
A chocolate factory reports that the average number of chocolates a person eats per year is normally distributed with a mean of 500 with a standard deviation of 20. What is the probability that a sample of 100 people will eat an average of at least 503 chocolates in a year?

Solution:
When solving any problem, first look at what the problem tells you:
mean = 500
std dev = 20
n = 100
The data is normally distributed (bell shaped)

In other words, we are told that we know that the mean chocolates eaten is 500 and the standard deviation is 20. We also know that the sample size in this question is n = 100.

Next, look closely at what the question is asking:
This question is asking for the probability that a sample of 100 people will eat an average of at least 503 chocolates in a year.
In short, the question is asking:
P( choc eaten by 100 people > 503)

How to solve problems like this:
Because we want the probability and because we know that our data is normally distributed, we will use the z Table to solve this question.

To use the z table, we must convert the values in the question to their equivalent z values using the formula:
z = (x – mean) / [stddev/sqrt(n)]
This formula will convert any value (call it x) into its equivalent z value.

In this problem:
We want P( choc eaten by 100 people > 503)
First, we must CONVERT 503 into its equivalent z value using the formula.

WARNING: You cannot use the simplified version of this formula which is z = (x – mean)/stddev. This simplified formula can only be used when your sample is of size one.

Convert 503 into a z value:
z = (503 – 500) / [20 / sqrt(100)]
z = (503 – 500) / [20 / 10]
z = (503 – 500) / 2
z = 3/2
z = 1.5

NOTE: We found the sqrt first. Then we did the division. Then subtract in the parenthesis. Finally the last division. You must solve these and all problems using the correct order of operations.

Next, now that we know that the z value for “503” (with a sample size of 100) is “1.5”, we can use the z table.

Recall that the question asked was:
P( choc eaten by 100 people > 503)

This is the same as
P( z > 1.5)

Using the z table, we must first get P(z < 1.5).
P(z < 1.5) = .9332 P(z > 1.5) = 1 – P(z < 1.5) = 1 – .9332 = .0668 or 6.68%

Final Solution:
Therefore, the probability that a sample of 100 people will eat an average of at least 503 chocolates in a year is about 6.68%

REMEMBER: The z-table ALWAYS gives you the probability of LESS THAN. If you need the probability to be greater than, you will subtract from 1.