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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​?

Respuesta :

JeanaShupp

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%

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