Expected Value of random variables using standard deviation? Question: The manager of a popular seafood restaurant estimates that the daily consumption of shrimp is normally distributed with a mean of 15 pounds and a standard deviation of 2.7 pounds. He makes it a point to buy the right amount of shrimp everyday to prevent waste and shortage. Calculate the amount of shrimp that should be bought daily so that it meets demand 92% of the days.
I'm under the impression to use the mean/expected value equation: ExP(x), but I'm not sure which values to place into the formula (my textbook doesnt use standard deviation in any of the examples). Or, is this question using binomial distribution probability? Can anyone help me better understand how to compute this? 
The answers are out of:
a.  12.44
b.  19.43
c.  18.93
d.  17.57
e.  10.57

 A: Look up a table of the normal distribution.  It will give you the number of standard deviations to get 92%.  It should be around 1.  Then add 2.7 times this standard deviation to 15 to get desired answer, about 18.
A: You have daily demand $X \sim \mathsf{Norm}(\mu =15, \sigma=2.7)$ and you seek $c$ such that $P(X \le c) = .92.$
In effect, you want the 92nd quantile of this distribution. Using R statistical software (or other
technology), the result is 18.794, which does not
exactly match any of your proposed answers.
qnorm(.92, 15, 2.7)
[1] 18.79369

I suppose the proposed answers account for slight
errors involved in standardizing and using printed
normal tables.
$$P(X \le c) = 
P\left(\frac{X-\mu}{\sigma} \le \frac{c-15}{2.7}\right) - 0.92.$$
The approximate 92nd percentile can be found correct
to a couple of decimal places using printed standard
normal CDF tables. I don't know what style of table
is provided in your text, but you look for the value
in the body of the table that comes closest to cutting off probability 0.08 from the upper tail of the
standard normal distribution. Then look in the margins of the table to find the approximate 92nd percentile.
The answer will be about 1.40 or 1.41. Let;s use 1.40 (as you did in your Comment),
qnorm(.92, 0, 1)
[1] 1.405072

Then $(c-15)/2.7 = 1.40$ and $c = 2.7(1.40)+15 = 18.78.$
If you use 1.41, the answer is 18.81. Perhaps you were supposed to use 92.5% in case 92% was rounded--as @herbsteinberg (+1) suggests. But how are you supposed to know that?
Simply put, either the question or the answer has a mistake. I have written a couple of books and errors invariably
slip in--no matter how careful one tries to be. In multiple choice questions (which I seldom
use), I have learned to say, "Which of the following
is closest to the correct answer"?
