Question

the placement test for a college has scores that are normally distributed with a mean of 400 and a standard deviation of 100. if the college accepts only the top 12% 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 12%, the area to the left of the cutoff score is \(1 - 0.12=0.88\). Looking up in the standard normal table (z - table), the z - score corresponding to an area of 0.88 is approximately \(z = 1.175\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 450\), \(\sigma = 100\), and \(z = 1.175\).
We want to solve for \(x\). Rearranging the formula gives \(x=\mu+z\sigma\).
Substitute the values: \(x = 450+1.175\times100\).

$$x=450 + 117.5$$

Answer:

\(567.5\)