Question

It is known that for all tests administered last​ year, the distribution of scores was approximately normal with mean 74 and standard deviation 7.1. a. A particular employer requires job candidates to score at least 80 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 80​?

Answer 1

Answer: 19.77%

Step-by-step explanation:

Given: Mean : $\mu=74$

Standard deviation : $\sigma = 7.1$

The formula to calculate z-score is given by :_

$z=\dfrac{x-\mu}{\sigma}$

For x= 80, we have

$z=\dfrac{80-74}{7.1}\approx0.85$

The P-value = $P(z>0.85)=1-P(z$

In percent , $0.1976626\times100=19.76626\%\approx19.77\%$

Hence, the approximate percentage of the test scores during the past year exceeded 80 =19.77%