In: Statistics and Probability
Weekly gross income earned by lecturers in XYZ University is normally distributed with a mean of $1,600 and a standard deviation of $250. The Chief Operation Manager of XYZ University would like to do an audit on the weekly gross income earned by lecturers in his institution. Assist the manager is answering this: what is the probability that if 50 lecturers are randomly selected, their average weekly gross income would be more than $1,700?
| a. | 
 0.9954  | 
|
| b. | 
 0.9977  | 
|
| c. | 
 0.0023  | 
|
| d. | 
 0.4977  | 
|
| e. | 
 0.5023  | 
Given that,
mean = 
 = 1600
standard deviation = 
 = 250
n=50

= 
=1600

= 
 / 
n = 250/ 
50 = 35.355
P(
 >1700 ) = 1 - P(
< 1700)
= 1 - P[(
- 
) / 
< (1700-1600) /35.355 ]
= 1 - P(z <2.83 )
Using z table
= 1 - 0.9977
= 0.0023