Question

the placement test for a college has scores that are normally distributed with a mean of 450 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 corresponding to the lower 88%

Since the college accepts the top 12%, we need the z - score for the lower \(100 - 12=88\%\) (or 0.88) of the data. Looking at the standard normal distribution table (z - score table), the z - score corresponding to a cumulative probability of 0.88 is approximately \(z = 1.175\).

Step2: Use the z - score formula \(z=\frac{x-\mu}{\sigma}\) to find \(x\)

We know that \(\mu = 450\), \(\sigma=100\), and \(z = 1.175\).
From \(z=\frac{x-\mu}{\sigma}\), we can solve for \(x\):

$$x=\mu+z\sigma$$

Substitute the values:

$$x = 450+1.175\times100$$

$$x=450 + 117.5$$

Answer:

\(x = 567.5\)