Hypergeometric approximated Normal/Gaussian distribtuion A large lecture theater has 270 seats, 24 of which can accomodate left-handed students.  Suppose it is known that 14% of people are left-handed.  One class held in the theater has 205 students.
(a) Let X be the number of left-handed students in the class.  The mean and standard deviation of X are, respectively,    and   
(b)  Let p hat be the proportion of left-handed students in the class.  The mean and standard deviation of p hat are, respectively,    and    .
(c)  What is the approximate probability that all left-handed students in the class get an appropriate desk?
I am assuming that this is supposed to be a hypergeometric probability that needs to be approximated to a normal distribution
mean is equation to 0.14*205 = 28.7
standard deviation is sqrt(205*.14*.86*((270-205)/(270-1)))=2.442140873
p hat = X/n
Not sure if that is even right. 
 A: It's necessary to make certain assumptions. 
If we assume that no right-handed person sits in a seat that can accommodate a left-hander, and we assume that people are only left- or right-handed (no ambidextrousness allowed), and we assume that the people in the class are effectively randomly selected with respect to handedness (e.g. no dependence due to being related or whatever) then the question reduces to:
"Given that 14% of people are left handed, out of 205 randomly selected people, what's the probability that no more than 24 will be left handed?"

Edit: explaining part c

(c) What is the approximate probability that all left-handed students in the class get an appropriate desk?

which I translated to:

"Given that 14% of people are left handed, out of 205 randomly selected people, what's the probability that no more than 24 will be left handed?"

Let $X$ be defined (as in your Q) as the number of left handed students in the class.
Under the assumptions we considered, $X$ is distributed $\text{binomial}(205,0.14)$, which is approximately $N(205 \times 0.14, 205 \times 0.14 \times (1-0.14))$ (i.e. $N(28.7,24.682)$).
$P(X\leq 24) = P(\frac{X-28.7}{\sqrt{24.682}}
\leq \frac{24-28.7}{\sqrt{24.682}})\approx P(\frac{X-28.7}{\sqrt{24.682}}
\leq -0.946) $
i. Ignoring continuity correction:
$P(X\leq 24) = P(\frac{X-28.7}{\sqrt{24.682}}
\leq \frac{24-28.7}{\sqrt{24.682}})\approx P(\frac{X-28.7}{\sqrt{24.682}}
\leq -0.946) \approx P(Z\leq -0.946) $
$\hspace{2.3cm}\approx 0.172$
ii. With continuity correction:

$P(X\leq 24) \approx P(\frac{X-28.7}{\sqrt{24.682}}
\leq \frac{24+0.5-28.7}{\sqrt{24.682}})\approx P(Z\leq -0.845...) $
$\hspace{2.3cm}\approx 0.199$
iii. Exact binomial answer: $0.20081886...$
So it's around 20% chance.
