Question

the placement test for a college has scores that are normally distributed with a mean of 600 and a standard deviation of 100. if the college accepts only the top 14% 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 corresponding to the percentile

The college accepts the top 14% of examinees. So we need to find the z - score corresponding to the percentile $100 - 14=86$th percentile. Looking up in the standard normal distribution table (z - score table), the z - score corresponding to a cumulative probability of 0.86 is approximately $z = 1.08$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the original normal distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know that $\mu = 600$, $\sigma=100$, and $z = 1.08$. We want to solve for $x$. Rearranging the formula gives $x=\mu+z\sigma$.

Step3: Calculate the cutoff score

Substitute the values into the formula: $x = 600+1.08\times100$.
$x=600 + 108=708$.

Answer:

708