Consider the following problem:
WINK, Inc. is made up of 450 employees who work a total of 13,500
hours per week. If the number of weekly work hours per person has a
normal distribution and the standard deviation equals 7 hours, how
many employees work more than 37 hours per week?
Although I have solved this question, it was very difficult, even using the Z score and Z table. Is there some trick that can be done here to solve this or similar questions?
I believe that there must be some trick, since in word problems you are not given a statistical Z table of the normal distribution.
Edit:
I have found the following trick, but I do not know what is happening here. But it gets the right answer. If somebody has time, I appreciate if they could explain to me how it is done?
Based on the given information, we know that
the average is 13,500 รท 450, or 30 hours. One standard deviation above the
mean is 37 hours. The bell curve of a normal distribution puts the mean
at 50%. One standard deviation in either direction is 34% away from the
mean. So, one standard deviation is 84% of the employees. Since 84% of
the employees work 37 hours, 16% work more than 37 hours. 16% of 450
employees = 72.
Best Answer
If the total number of hours worked $N=13500$, and the number of employees is $n=450$, then the mean number of hours worked per employee is given by:
$$\mu = \frac{N}{n}=30\:\text{hours}$$
We are given that the standard deviation $\sigma = 7\:\text{hours}$, and therefore we have that the number of hours worked per employee is normally distributed $T \sim \mathcal{N}(30, 49)$. We want to find the following probability:
$$P(T\gt 37) = 1-P(T\leq 37) = 1-\Phi\left(\frac{37-30}{7}\right) = 0.158655$$
However, a "trick" way of doing this would be to remember that approximately $68\%$ of the normal distribution lies within $1\sigma$ deviation, and thus we have that:
$$P(T > \mu + \sigma)\approx 1-(50\%+34\%)=0.16$$