Normal Distribution Probability 
*

*At a large publishing company, the mean age of proofreaders is 36.2 years, and the
standard deviation is 3.7 years. Assume the variables are normally distributed.


a. If a proofreader in the company is randomly selected, find the probability that
his or her age will be between 36 and 37.5 years
Attempt (unverified): 
Probability distribution of right-hand side of the mean on the bell curve:
35% of 1 Standard Deviation = 1.295 years
36.2+1.295 = 37.495 (approx. 37.5 years)
Therefore, 35% of 34% (half of one standard deviation on the right) = 11.9%
We do the same thing for the left-hand side and end up with: 6.25% of 34% (half of 1s to the left) = 2.125% [gives us 35.97 years =~ 36 years]
Therefore, P(age between 36 and 37.5 years) = 11.9 + 2.125 = 14.025% (final answer)
b. If a random sample of 15 proofreaders is selected, find the probability that the
mean age of the proofreaders in the sample will be between 36 years and 37.5
years.
P(age between 36 and 37.5) = 14.025% [from A]
n = 15
-=HOW DO I PROCEED FROM HERE=- ? 
PLEASE ASSIST. Thank you. 
 A: I do not understand the first calculation. We want the probability that the age is between $\frac{0.2}{3.7}$ standard deviation units below the mean and $\frac{1.3}{3.7}$ standard deviation units above the mean. 
Using the table of the standard normal, we find that the probability that the age is  $\frac{1.3}{3.7}\approx 0.351$ standard units above the mean or less is approximately $0.637$. 
The probability that we are below the mean by $\frac{0.2}{3.7}$ or more standard deviation units is about $0.48$. So our required probability is about $0.637-0.48$, some distance from your number. 
For the next problem, if we take $15$ ages and average them, the resulting random variable has normal distribution with mean $36.2$ and standard deviation $\frac{3.7}{\sqrt{15}}\approx 0.955$. Now one needs to essentially repeat the first calculation, but with $0.955$ replacing $3.7$. The probabilities will change dramatically.
Added: Let $X$ be normally distributed with mean $\mu$ and standard deviation $\sigma$. Then
$$\Pr(X\le a)=\Pr\left(Z\le \frac{a-\mu}{\sigma}\right),$$ 
where $Z$ is standard normal. 
For reasonable positive values of $z$, $\Pr(Z\le z)$ can be looked up approximately in a table of the standard normal.
For negative $z$, we have to work a bit. If $z$ is negative, then $\Pr(Z\le z)=1-\Pr(Z\le |z|)$.
Tables of the standard normal are available on the web. Typically they are a single page, you can download one and print it. Such a table is still often found at the back of introductory statistics books. 
But nowadays the reasonable thing is to let software do the work for you. There are many pieces of software, including spreadsheet programs, that have normal distribution calculations as a built in feature.
