Question

e of Scores, a publication of the Educational Testing Service, the scores on the verbal portion of the GRE have mean 150 points and standard deviation 8.75 points. Assuming that these scores are (approximately) normally distributed, a. obtain and interpret the quartiles. b. find and interpret the 99th percentile.

Answer 1

a) Q1= 144.10

Median = 150

Q3=155.90

b) The 99 percentile would be:
`a=150 +2.33*8.75=170.39`

And represent a value who accumulate 99% of the values below

Explanation:

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

`X \sim N(150,8.75)`

Where
`\mu=150` and
`\sigma=8.75`

Part a

Lets begin with the first quartile:

`P(X>a)=0.75` (a)

`P(X<a)=0.25` (b)

We can find the quantile in the normal standard distribution and we got z=-0.674.

And we can apply the z score formula and we got:

`z=-0.674<(a-150)/(8.75)`

And if we solve for a we got

`a=150 -0.674*8.75=144.10`

The median for this case is the mean
`Median =150`

For the third quartile we find the quantile who accumulate 0.75 of the area below and we got z=0.674 and we got:

`a=150 +0.674*8.75=155.90`

Part b

We can find the quantile in the normal standard distribution who accumulate 0.99 of the area below and we got z=2.33.

And we can apply the z score formula and we got:

`z=2.33<(a-150)/(8.75)`

And if we solve for a we got

`a=150 +2.33*8.75=170.39`

And represent a value who accumulate 99% of the values below