# A college placement office con

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

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

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

c) What would be the90^{th}percentile for the income of these individualengineers?

d) What would be the90^{th}percentile for the average income of groups of 9engineers?

e) Why is the 90^{th}percentile in (d) a smaller numberthan the value in (c)? Explain using the bell curve and whathappens to it when you are looking at the average of a group.

Answer:

**Solution:**

We are given:

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

**Answer:** Let x be the amount earned byengineers.

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 4engineers 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 4engineers will be between $55,040 and $88,960 is **

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

**Answer:** To find the 90th percentile for theincome of these individual engineers, we need to find the z valuecorresponding 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 individualengineers is

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

**Answer:** To find the 90th percentile for theaverage income of group of 9 engineers, we need to find the z valuecorresponding 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 9engineers is

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

**Answer:** The 90thpercentile in (d) is smallernumber than the value in (c) because the standard deviation of thesample mean decreases upon increasing the sample size. The samplesize in part c is 1, while the sample size is 9 in part (d). Thatis the reason we reason we see the 90th percentile in part (d) issmaller 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 thestandard gets reduced upon increasing the sample size