Question

the placement test for a college has scores that are normally distributed with a mean of 500 and a standard deviation of 80. if the college accepts only the top 16% of examinees, what is the cutoff score on the test for admission? click the icon to view the table of z - scores and percentiles. the cutoff score is

Explanation:

Step1: Find the z - score

Since the college accepts the top 16%, we need to find the z - score corresponding to the \(100 - 16=84\)th percentile. Looking at the standard normal distribution table (z - score table), the z - score corresponding to a cumulative probability of 0.84 is \(z = 1\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 500\) (mean), \(\sigma = 80\) (standard deviation), and \(z = 1\).
We want to solve for \(x\). Rearranging the formula gives \(x=\mu+z\sigma\).
Substitute the values: \(x = 500+1\times80\).
\(x=500 + 80\).

Answer:

\(580\)