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.
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
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