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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 493}{108}$

$X - 493 = 1.28*108$

$X = 631.24$

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

Answer 2

Answer:

$b=493 +1.28*108=631.24$

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

Step-by-step 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:

$P(X<b)=P(\frac{X-\mu}{\sigma}<\frac{b-\mu}{\sigma})=0.90$  

$P(z<\frac{b-\mu}{\sigma})=0.90$

$z=1.28<\frac{b-493}{108}$

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