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%

Explanation:

Given: Mean :
`\mu=74`

Standard deviation :
`\sigma = 7.1`

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

`z=(x-\mu)/(\sigma)`

For x= 80, we have

`z=(80-74)/(7.1)\approx0.85`

The P-value =
`P(z>0.85)=1-P(z<0.85)=1-0.8023374=0.1976626`

In percent ,
`0.1976626*100=19.76626\%\approx19.77\%`

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