Question

A college placement office conducted a survey of 100 engineers who had graduated from Stanford University. For these engineers, the mean salary was computed to be $72,000 with a standard deviation of $22,000.  The distribution of salary is roughly bell shaped.

a) What percentage of these engineers will earn between $55,040 and $88,960?

b) What is the probability that the average income of 4 engineers will be between $55,040 and $88,960?

c) What would be the 90^thpercentile for the income of these individual engineers?

d) What would be the 90^thpercentile for the average income of groups of 9 engineers?

e) Why is the 90^thpercentile in (d) a smaller number than the value in (c)? Explain using the bell curve and what happens to it when you are looking at the average of a group.

Answer 1

Solution:

We are given:

a) What percentage of these engineers will earn between $55,040 and $88,960?

Answer: Let x be the amount earned by engineers.

Therefore, we have to find

When , we have:

  

  

  

Now, when , then we have:

  

  

  

Therefore, we have to find . Using the standard normal table, we have:

or 55.88%

Therefore, 55.88% of these engineers earn between $55,040 and $88,960

b) What is the probability that the average income of 4 engineers will be between $55,040 and $88,960?

Answer: Let be the average income of 4 engineers.

Therefore, we have to find

When , we have:

  

  

  

Now, when , then we have:

  

  

  

Therefore, we have to find . Using the standard normal table, we have:

Therefore,the probability that the average income of 4 engineers will be between $55,040 and $88,960 is

c) What would be the 90th percentile for the income of these individual engineers?

Answer: To find the 90th percentile for the income of these individual engineers, we need to find the z value corresponding to probability 0.90. And same is given below:

Now using the z score formula, we have:

Therefore, 90th percentile for the income of these individual engineers is

d) What would be the 90thpercentile for the average income of groups of 9 engineers?

Answer: To find the 90th percentile for the average income of group of 9 engineers, we need to find the z value corresponding to probability 0.90. And same is given below:

Now using the z score formula, we have:

Therefore, 90th percentile for the average income of group of 9 engineers is

e) Why is the 90thpercentile in (d) a smaller number than the value in (c)? Explain using the bell curve and what happens to it when you are looking at the average of a group.

Answer: The 90thpercentile in (d) is smaller number than the value in (c) because the standard deviation of the sample mean decreases upon increasing the sample size. The sample size in part c is 1, while the sample size is 9 in part (d). That is the reason we reason we see the 90th percentile in part (d) is smaller than the value in part (c).

We can show this using the bell curve:

From the above bell shaped curves, we can clearly see how the standard gets reduced upon increasing the sample size