There are 45 students in an elementary class... There are 45 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen examination paper is a random variable with an expected value of 5 min and a standard deviation of 4 min
a) if grading times are independent and instructor begins grading at 6:50 PM and grade continuously, what is the approximate probability that he is through grading before the 11:00 PM TV news begin?
for this part, I'm not sure how to approach, the thing that really throws me off is finding the probability between 6:50 through 11:00. am I supposed to find the number of minutes between these two hours? and find the probability of that?
 A: As the number of papers gets large the distribution of the average time spent grading a paper increasingly resembles a normal distribution.  (Central Limit Theorem)  At $n=45$ it is reasonable to use the normal distribution with ($\mu = 5, \sigma = 4$).
The sum of $n$ normally distributed random variables is $n\mu$
and standard deviation is $\sqrt n \sigma$  
A: For the grading time of an individual paper $X_i$ you have population mean $\mu_X = 5$ and 
population standard deviation $\sigma_X = 4.$
For the total grading time $T$ of $n = 45$ papers you have population mean $\mu_T = n\mu_x$ and population variance and $\sigma_T^2 = n\sigma_X^2.$ 
You should find an explanation for means and variances of totals in your
textbook to verify this.
Although grading times of individual papers are not said to be normal,
you can hope that, for a total of $n=45$ papers, the Central Limit Theorem
has the effect of making $T$ approximately normal.
There are $4 \times 60 + 10 = 250$ minutes of grading time before the TV news
begins. So you want $P(T < 250).$ You have not shown any work to give a
clue whether you are expected to use printed normal tables or to use
software to compute that probability. You should look for examples
in your textbook for that. (Perhaps you need to standardize to get a z-score.)
Using software I get about 0.82.
Now it's your turn to do some work on this.
