Question

a banks loan officer rates applicants for credit. the ratings are normally distributed with a mean of 200 and a standard deviation of 50. find ( p_{60} ), the score which separates the lower 60% from the top 40%. round to one decimal place.

a. 211.3
b. 212.7
c. 207.8
d. 187.5

Explanation:

Step1: Find the z - score

We want to find the \(z\) - score corresponding to an area of \(0.60\) (lower \(60\%\)) in the standard normal distribution table. Looking up in the standard normal table (or using a calculator with a normal - distribution function), the \(z\) - score \(z\) such that \(P(Z\lt z)=0.60\) is approximately \(z = 0.25\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 200\) (mean), \(\sigma=50\) (standard deviation), and \(z\) is the z - score from Step 1.
We need to solve for \(x\). Rearranging the formula gives \(x=\mu + z\sigma\).
Substitute \(\mu = 200\), \(z = 0.25\), and \(\sigma = 50\) into the formula:
\(x=200+0.25\times50\)
\(x=200 + 12.5\)
\(x=212.5\)

Another way is to use a more accurate \(z\) - value. Using a calculator (e.g., TI - 84: invNorm(0.60,200,50)), we get a more accurate \(z\) - value.
The inverse - normal function \(invNorm(p,\mu,\sigma)\) where \(p = 0.60\), \(\mu=200\), \(\sigma = 50\)
\(x=invNorm(0.60,200,50)\approx212.7\)

Answer:

B. 212.7