Q:

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​?

Accepted Solution

A:
Answer: 19.77%Step-by-step explanation:Given: Mean : [tex]\mu=74[/tex]Standard deviation : [tex]\sigma = 7.1[/tex]The formula to calculate z-score is given by :_[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x= 80, we have[tex]z=\dfrac{80-74}{7.1}\approx0.85[/tex]The P-value = [tex]P(z>0.85)=1-P(z<0.85)=1-0.8023374=0.1976626[/tex]In percent , [tex]0.1976626\times100=19.76626\%\approx19.77\%[/tex]Hence, the approximate percentage of the test scores during the past year exceeded 80 =19.77%