Question

According to a company’s website, the top 10% of the candidates who take the entrance test will be called for an interview. The reported mean and standard deviation of the test scores are 65 and 9, respectively. If test scores are normally distributed, what is the minimum score required for an interview? (You may find it useful to reference the z table. Round your final answer to 2 decimal places.)

Answer 1

To find the minimum score required for an interview, we need to determine the cutoff score that corresponds to the top 10% of test scores.

First, we need to find the z-score corresponding to the top 10%. The z-score can be found using the z-table.

The z-score can be calculated using the formula:
z = (x - μ) / σ

Where:
x = raw score
μ = mean
σ = standard deviation

Since we want the top 10%, we are looking for a z-score that corresponds to a cumulative probability of 0.90.

Using the z-table, we find that a z-score of 1.28 corresponds to a cumulative probability of 0.90.

Now we can solve for the minimum score required for an interview.

1.28 = (x - 65) / 9

Simplifying the equation:
1.28 * 9 = x - 65
11.52 = x - 65
x = 11.52 + 65
x = 76.52

Therefore, the minimum score required for an interview is 76.52 (rounded to 2 decimal places).