Question

the placement test for a college has scores that are normally distributed with a mean of 600 and a standard deviation of 90. if the college accepts only the top 10% 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 for the top 10%

The college accepts the top 10% of examinees, so the area to the left of the cutoff z - score is $1 - 0.10=0.90$. Looking up the z - score in the standard normal distribution table (or using a calculator/statistical software) for an area of 0.90, we find that the z - score ($z$) is approximately 1.28 (this value is obtained from the standard normal table where $P(Z\leq z)=0.90$).

Step2: Use the z - score formula to find the cutoff score ($x$)

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value we want to find. We know that $\mu = 600$, $\sigma=90$, and $z = 1.28$. Rearranging the formula to solve for $x$ gives $x=\mu+z\sigma$.

Substitute the known values into the formula:
$x = 600+1.28\times90$

First, calculate $1.28\times90 = 115.2$

Then, add this to the mean: $x=600 + 115.2=715.2$

Answer:

715.2