Normal distribution with sample

Question

I'm trying to figure out the best approach to this problem. I would assume that I can use the Central Limit theorem first and then a binomial cdf:

Chocolate is packaged into jars using a computerized system. The volume of one candy that can fill one jar follows a normal approximation with a mean of 300 ml and $\sigma$ = 24ml. In an experiment, 15 random jars were selected. What was the probability that there were at most 2 jars that had less than 275 ml of candy in its jar?

$P(\frac{X-300}{24}<\frac{275-300}{24}) = P(Z<-1.04) = .1429$

Then

$P(Y \leq 2) = 105(.1429)^{2}(.8571)^{13}+15(.1429)(.8571)^{14}+(.8571)^{15} = 0.635$

Not sure if this is the right approach or not. The answer isn't correct, just wondering where I am off.

Answer 1

Hint: $P(Z<-1.04)=0.14\color{red}{92}$. The last two digits has to be the other way round. Therefore the calculation is

$$105(.1492)^2 \cdot(.8508)^{13}+15\cdot(.1492)(.8508)^{14}+(.8508)^{15}=0.61$$

Answer 2

$\dbinom{15}2 = \dfrac{15\times14}{2\times1} = 105$, so you need $105$ where you have $150$.

Other than that your answer looks OK.

(I doubt anyone would really tolerate a standard deviation that big, but I don't think this was intended to be realistic.)