| The Problem |
|
The average hourly wage of fast-food workers employed by a nationwide chain is $5.55. The standard deviation is $1.15. If a sample of 50 workers is selected, find the probability that the mean of the sample will be between $5.25 and $5.90. |
Since this problem
deals with a distribution of averages, the Cental Limit Theorem applies.
Thus, the distribution is normal with standard deviation of 
.
The TI84 has a function
that preforms the required calculation. That function is
From the statement
of the problem, we have 
= 5.55, 
=
1.15, n = 50, 
=
5.25, and 
=
5.90.
From the home screen,
press 2nd Vars[DISTR] and then press 2 to select 2:normalcdf(.
Press ENTER. Type 5.25 ENTER 5.90 ENTER 5.55 ENTER 
then press ENTER twice.
The results are shown below.
 |
 |
Thus, the probability that the mean of the sample will be between $5.25 and $5.90 is about 95%.
|
|
|
|
© 2019, Middle Georgia State University. All rights reserved.
