Question

Suppose SAT Writing scores are normally distributed with a mean of 493 and a standard deviation of 108. A university plans to send letters of recognition to students whose scores are in the top 10%. What is the minimum score required for a letter of recognition

Answer 1

`b=493 +1.28*108=631.24`

The minimum score required for a letter of recognition would be 631.24

Explanation:

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

`X \sim N(493,108)`

Where
`\mu=493` and
`\sigma=108`

On this questio we want to find a value b, such that we satisfy this condition:

`P(X>b)=0.10` (a)

`P(X<b)=0.90` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find b.

As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

`P(X<b)=P((X-\mu)/(\sigma)<(b-\mu)/(\sigma))=0.90`

`P(z<(b-\mu)/(\sigma))=0.90`

`z=1.28<(b-493)/(108)`

And if we solve for a we got

`b=493 +1.28*108=631.24`

The minimum score required for a letter of recognition would be 631.24

Answer 2

The minimum score required for a letter of recognition is 631.24.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 493, \sigma = 108`

What is the minimum score required for a letter of recognition

100 - 10 = 90th percentile, which is the value of X when Z has a pvalue of 0.9. So X when Z = 1.28.

`Z = (X - \mu)/(\sigma)`

`1.28 = (X - 493)/(108)`

`X - 493 = 1.28*108`

`X = 631.24`

The minimum score required for a letter of recognition is 631.24.