It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 74 and sta

Question

3 years ago

14

It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 74 and sta

ndard 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?

Mathematics

1 answer:

Lunna [17] · Answer 1 · 2022-08-10T06:42:53+03:00

Lunna [17]3 years ago

6 0

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%

It is known that for all tests administered last​ year, the distribution of scores was approximately normal with mean 74 and sta

It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 74 and sta