Question

Scores on a test have a mean of 71 and Q3 is 82. The scores have a distribution that is approximately normal. Find P90

Answer 1

Given:

Mean = 71

Q3 = 75th percentile = 82

We also know that

From the standard normal table or using Excel the corresponding z-value for 0.75(Q3 or 75th percentile) is calculated as 0.6745. (In Excel, use the function NORMSINV(0.75))

Therefore, the above equation of Z can be re-written for the value of 82 (Q3) as

Now, solving the above equation we have

Now, we have found standard deviation .

From the standard normal table or using Excel we can find the Z-value corresponding to 90th percentile as 1.2816 (In Excel, use the function NORMSINV(0.90)).

Hence, 90th percentile value (value of x in the Z formula) is calculated using the above z-formula as below

Hence, P90 = 92 (approx).