# 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 the90thpercentile for the income of these individualengineers?

d) What would be the90thpercentile for the average income of groups of 9engineers?

e) Why is the 90thpercentile 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.

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

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