All tutors that are employed at a local college must take a test and must score in the top 20%. if the scores are normally distributed with a mean of 78 and a standard deviation of 5, what is the cutoff score to be hired?

To solve this problem, we use the z statistic with formula:

z = (x – u) / s

or

x = s z + u

First we find the value of z at P = top 20%

z = 0.84

So the cut-off is:

x = 5 * 0.84 + 78

x = 82.2                 (ANSWER)

step-by-step explanation:

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

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