Continuous random variable distribution probability In a final exam, there are 300 students and one-third of them bring a water bottle. Suppose that for those who bring a water bottle, the chance that each of them leaves behind his/her water bottle in the examination room after the exam is 0.06. 
a) What are the mean and standard deviation of the total number of water bottles left behind after the exam? Give your answers to 2 decimal places. 
b) Using an appropriate approximation method, find the probability that 5 or less water bottles are left behind. Carry your answers from the previous part to at least 6 decimal places in your calculation here. Give you final answer to 4 decimal places. 
c) Using an appropriate approximation method, at most how many water bottles will be left behind 84.13% of the time? Give your answer to the nearest integer.
So far, I have found the mean. I think it should be 300$\frac{1}{3}$(0.06) = 6. I got the standard deviation as 23.74 by doing $\sqrt{Var(X)}$ where Var(X)= E($X^{2}$) - $[E(X)]^{2}$. I'm pretty lost because I don't know how to determine which distribution to use. Any hints or help is appreciated.
 A: I suppose they want you to find the binomial distribution of the
number $X$ of bottles left behind. That's $X \sim \mathsf{Binom}(n=100,p=.06)$.
Then $\mu = E(X) = np = 6$ as you say, and $\sigma = SD(X) = \sqrt{np(1-p)} = 2.37.$
For part (b), you want 
$$P(X \le 5) = P(X \le 5.5) = P\left(\frac{X - \mu}{\sigma} \le 
\frac{5.5 - 6}{2.37}\right) \approx P(Z \le -0.21) =\, ??,$$
where $Z$ is standard normal. You can get the probability from printed
tables of the standard normal CDF. Using 5.5 instead of 5.0 is called
the 'continuity correction' for the normal approximation.
The exact binomial probability can be found using software. Here is how
that looks using R statistical software. You should get close to 0.44
using the normal approximation.
pbinom(5, 100, .06)
## 0.4406927

Here is a plot of the PDF of $\mathsf{Binom}(100,.06)$ along with the
density function of the 'best fitting' normal distribution. (As @Callculus anticipates, the fit
isn't perfect, so don't expect to get exactly 0.44.)

In part (c), you want to find $k$ such that $P(X \le k) = 0.8413.$
I will leave that computation to you. After you 'standardize' you
will need to use the normal table 'backwards', looking in the body
of the table for 0.8413 (or 0.3413) depending on the style of
normal table you are using. Then find the corresponding z-value from the margins of the table, and
solve for $k.$ Just by eye from the plot above, I'd guess $k \approx 8.$ 
