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