|
Question 1182643: Business Weekly conducted a survey of graduates from 30 top MBA programs.
On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is $187,000.
Assume the standard deviation is $45,000.
Suppose you take a simple random sample of 36 graduates.
Round to 4 decimal places.
What is the distribution of X?
For a single randomly selected graduate, find the probability that her salary is between $186,650 and $195,500.
For the simple random sample of 36 graduates, find the probability that the average salary is between $186,650 and $195,500.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The distribution of X is approximately normal with mean $187,000 and variance 45000^2 $^2.
a. z=(x bar-mean)/sigma
>=-350/45000 which is >=-0.00778
<=(8500)/45000 or 0.18888
that probability is 0.0780 (unrounded, which z often is not, would be 0.0771
b .for 36 people the denominator would be 45000/sqrt(36) or 7500
This probability would be 0.3901 (not rounded).
The point of this question is that it would be quite unlikely for a single person to have a salary in this restricted area around the mean, but the mean of a group of 36 would be far more likely to cluster around the mean.
|
|
|
| |
