Question

Ron thinks he's kind of a big deal and decides to challenge Veronica to an Anchor-person Aptitude Test (AAT). The average anchor person scores 100 points (out of 200 possible points), with a standard deviation of 4.5 points. Brick, the newsroom shuttle diplomat, gets both parties to agree that being a "big deal" would require an anchor person's score to beat 97.5% of the other scores. Ron scores 111 and Veronica scores 117. Assuming SAT scores are normally distributed, which of the following statement is true
a) Only Ron is a Big Deal.
b) Both Veronica and Ron are a Big Deal.
c) Only Veronica is a Big Deal.
d) Neither Veronica nor Ron are a Big Deal

Answer 1

Answer:

b) Both Veronica and Ron are a Big Deal.

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:

$\mu = 100, \sigma = 4.5$

Big deal:

Beat 97.5% of the other scores, which means that Z must have a pvalue of at least 0.975.

Ron

Scored 111, so $X = 111$

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

$Z = \frac{111 - 100}{4.5}$

$Z = 2.44$

$Z = 2.44$ has a pvalue of 0.9927, so Ron is a big deal

Veronica

Scored 117, so $X = 117$

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

$Z = \frac{117 - 100}{4.5}$

$Z = 3.78$

$Z = 3.78$ has a pvalue of 1, so Veronica is a big deal

So the correct answer is:

b) Both Veronica and Ron are a Big Deal.