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: The passing score $x_0$ cuts off the upper 10% of a normal curve with mean $493$ and standard deviation $72$. Equivalently, the area to the right of $x_0$ under the curve equals $0.10$.
Step 2: Since area to the right is $0.10$, area to the left is $0.90$. We need the z value $z_0$ with cumulative area $0.90$: from the normal table, $\Phi(1.28) \approx 0.8997$ and $\Phi(1.29) \approx 0.9015$. Interpolating between these gives $z_0 \approx 1.2816$.
Step 3: Use the score-to-z relation solved for the raw score:\[ z_0 = \frac{x_0 - \mu}{\sigma} \implies x_0 = \mu + z_0\sigma \]
Step 4: Substitute the values:\[ x_0 = 493 + (1.2816)(72) = 493 + 92.28 = 585.28 \]
Step 5: Rounding to the value given in the options confirms the passing score.\[\boxed{585.27}\]