Suppose that scores on a test are normally distributed with a mean of 80 and a standard deviation of 8...

Question

a) What is the 90^th Percentile?

b) What percentage of students score less than 90?

Questions given in my Stat Methods course. I had easily finished my homework over Z-scores, density curves, etc but can't solve this question for the life of me. Maybe I'm making it seem harder than it actually is?

Answer 1

The 90th percentile is the score that 90% of the testers score below. So, on the normal table, you would find the area that is closest to .9, which is 1.28. That is the z-score. So then we can convert it to a regular score by using the z-score formula: z = (x - μ) / σ.

1.28 = (x - 80) / 8

10.24 = x - 80

90.24 = x

This can also be done using invNorm(0.9, 80, 8) on the TI-84, and it gives you 90.2524, which is really close to the other figure.

The percentage of scores less than 90 is just P(X < 90). We can use the z-score formula and the normal table as follows:

z = (90 - 80) / 8 = 1.25

The area for this z-score is 0.89435 , or 89.435%.

This can also be done using normalcdf on the TI-84, with upper 90, μ 80, and σ 8. This yields the same result of 0.89435 (with a few extra decimal places).