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 ÷ 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.
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:
$$P(T > \mu + \sigma)\approx 1-(50\%+34\%)=0.16$$