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 1% 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 z - score

The college accepts the top 1% of examinees. So the area to the left of the cutoff z - score in the standard normal distribution is $1 - 0.01=0.99$. Looking up this value in the z - score table, the z - score corresponding to an area of 0.99 is approximately $z = 2.33$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the raw score, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know that $\mu = 400$, $\sigma = 100$, and $z = 2.33$. Rearranging the formula for $x$ gives $x=\mu+z\sigma$.

Step3: Calculate the cutoff score

Substitute the values into the formula: $x = 400+2.33\times100$.
$x=400 + 233=633$.

Answer:

633