What's the mean median and mode of 8,2,3,5,7
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Answer:
The best choice would be hiring a random employee from company A
Step-by-step explanation:
<em>Supposing that the performance rating of employees follow approximately a normal distribution on both companies</em>, 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
where
= mean of company A
= 5.5 (average of rating between 1 and 10)
s = standard deviation of company A
We do the same for company C
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