Question

The board of examiners that administers the real estate broker's examination in a certain state found that the mean score on the test was 493 and the standard deviation was 72. If the board wants to set the passing score so that only the best 10% of all applicants pass, what is the passing score? Assume that the scores are normally distributed.

• A. 400.73
• B. 585.27
• C. 550.75
• D. 425.12

Hint:

Find the z-score corresponding to the 90th percentile (since top 10% pass), then convert back using \(x=\mu+z\sigma\).

Answer 1

The Correct Option is B

Step 1: Let \(X\) be the test score, \(X \sim N(493, 72^2)\). We want the score \(x_0\) such that only the top 10% score above it, i.e. \(P(X \ge x_0) = 0.10\), equivalently \(P(X \le x_0) = 0.90\).
Step 2: Standardize: \(Z = \dfrac{X - 493}{72}\). We need \(z_0\) such that \(P(Z \le z_0) = 0.90\).
Step 3: From the standard normal table, the 90th percentile is \(z_0 = 1.28\) (more precisely 1.2816).
Step 4: Convert back to the raw score:\[ x_0 = \mu + z_0 \sigma = 493 + 1.2816 \times 72 \]
Step 5: Compute: \(1.2816 \times 72 = 92.28\), so\[ x_0 = 493 + 92.28 = 585.28 \]This matches option (B) 585.27 (rounding of \(z_0\)). So the passing score is \(\boxed{585.27}\), option (B).