Answer:
The best choice would be hiring a random employee from company A
Step-by-step explanation:
Supposing that the performance rating of employees follow approximately a normal distribution on both companies, we are interested in finding what percentage of employees of each company have a performance rating greater than 5.5 (which is the mean of the scale), when we measure them in terms of z-scores.
In order to do that we standardize the scores of both companies with respect to the mean 5.5 of ratings
The z-value corresponding to company A is
[tex]z=\frac{\bar x-\mu}{s}[/tex]
where
[tex]\bar x[/tex] = mean of company A
[tex]\mu [/tex] = 5.5 (average of rating between 1 and 10)
s = standard deviation of company A
[tex]z=\frac{\bar x-\mu}{s}=\frac{6.5-5.5}{2.1}=0.7142[/tex]
We do the same for company C
[tex]z=\frac{\bar x-\mu}{s}=\frac{7.4-5.5}{6.8}=0.2794[/tex]
This means that 27.49% of employees of company C have a performance rating > 5.5, whereas 71.42% of employees of company B have a performance rating > 5.5.
So, the best choice would be hiring a random employee from company A
