Question

According to a company’s website, the top 25% of the candidates who take the entrance test will be called for an interview. You have just been called for an interview. The reported mean and standard deviation of the test scores are 68 and 8, respectively. If test scores are normally distributed, what is the minimum score required for an interview?

Answer 1

Answer:

The minimum score required for an interview is 73.4.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 68.8, \sigma = 8$

If test scores are normally distributed, what is the minimum score required for an interview?

Top 25%, which is at least the 100-25 = 75th percentile, which is the value of X when Z has a pvalue of 0.75. So it is X when Z = 0.675.

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 68}{8}$

$X - 68 = 0.675*8$

$X = 73.4$

The minimum score required for an interview is 73.4.