On a test at Microsoft, which has a mean of 500 and a standard deviation of 100, only the top 2% of the test takers will get an interview with Bill Gates. What score or better will they have to earn on this test to get the interview?
Please show the work. I can use excel so if you could explain how to do it in there, that would be even better! :)
Assume Normal distribution, top 2% corresponds to 100 - 2 = 98th percentile.
Z value for 98th percentile is 2.0537
Score to get the interview Mean + Z * Std deviation
= 500 + 2.0537 * 100
= 705.37
Using Excel, use the formula, to get the answer,
= 500 + NORMSINV(0.98) * 100
Here, NORMSINV(0.98) calculates Z value for 98th percentile
On a test at Microsoft, which has a mean of 500 and a standard deviation of...
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