Question

the placement test for a college has scores that are normally distributed with a mean of 400 and a standard deviation of 60. 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: Determine the percentile

The college accepts the top 12% of examinees. So we need to find the score corresponding to the 88th percentile (100 - 12 = 88).

Step2: Find the z - score

Looking up the 88th percentile in the z - score table, the z - score $z$ is approximately 1.175.

Step3: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value we want to find, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know $\mu = 400$, $\sigma=60$, and $z = 1.175$. Rearranging the formula for $x$ gives $x=\mu+z\sigma$.
Substitute the values: $x = 400+1.175\times60$.
$x=400 + 70.5$.
$x = 470.5$.

Answer:

470.5