Question

Scores for a civil service exam are normally distributed with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment?

Answer 1

Answer:

The lowest score you can earn and still be eligible for employment is 85.6925

Step-by-step explanation:

Problems of normally distributed samples can be 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 question, we have that:

$\mu = 75, \sigma = 6.5$

Top 5%.

At least the 100-5 = 95th percentile.

The 95th percentile is X when Z has a pvalue of 0.95. So X when Z = 1.645.

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

$1.645 = \frac{X - 75}{6.5}$

$X - 75 = 1.645*6.5$

$X = 85.6925$

The lowest score you can earn and still be eligible for employment is 85.6925

Answer 2

Answer:

The lowest score you can earn and still be eligible for employment is 85.70.

Step-by-step explanation:

We are given that Scores for a civil service exam are normally distributed with a mean of 75 and a standard deviation of 6.5.

Also, to be eligible for civil service employment, you must score in the top 5%.

Let X = Scores for a civil service exam

SO, X ~ N($\mu = 75,\sigma^{2} = 6.5^{2}$)

The z-score probability distribution is given by ;

                  Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean score = 75

            $\sigma$ = standard deviation = 6.5

Now, the lowest score that we can earn and still be eligible for employment is given by ;

              P(X $\geq$ $x$ ) = 0.05   {where $x$ is minimum score required}

                                               to be in top 5%}

             P( $\frac{X-\mu}{\sigma}$ $\geq \frac{x-75}{6.5}$ ) = 0.05

             P(Z $\geq \frac{x-75}{6.5}$ ) = 0.05

Now, in z table we will find out that critical value of X for which the area is in top 5%, which comes out to be 1.6449.

This means;         $\frac{x-75}{6.5} = 1.6449$

                          $x-75=1.6449 \times 6.5$  

                              $x$ = 75 + 10.69185 = 85.70

Therefore, the lowest score that we can earn and still be eligible for employment is 85.70.