A store sells two different coffee makers: a basic model for \$30, and a fancier model for \$50. We assume that different buyers’ choices don’t affect each other, and that different buyers share the same (constant) probability for whether they choose the cheap or the fancy model. Because of effective marketing, the store estimates that customers have a 65% chance of choosing the expensive coffee maker over the cheap one. Let $X$ be the number of people who opted for the fancy coffee maker, and let $h(X)$ represent the store's revenue. If the store sells 100 coffee makers on a given day, use the central limit theorem to calculate the approximate probability that the total revenue will be higher than \$4500.
I found $E[h(x)]=4300$ and that $SD[h(X)]=10\sqrt{91}$, but I'm not sure how to apply the CLT to find this probability. Below is what I've found so far.
$$P(h(X)>45000)=1-P(h(X)\leq4500=P(X\leq75)$$
I tried to use R to calculate this, but I'm not entirely sure if I'm on the right path, and therefore don't know what to put in the program. Any help (preferably coding assistance) would be very appreciated!
$h(X) = 50X + 30(100-X) = 20X + 3000$, so "revenue higher than 4500" corresponds to $X > 75$ as you have already found.
$X$ follows a $\text{Binomial}(100, 0.65)$ distribution. Compute the mean $\mu$ and variance $\sigma^2$ of this distribution. By the central limit theorem (or the normal approximation to the binomial), this binomial distribution can be approximated by a $\mathcal{N}(\mu, \sigma^2)$ distribution. Then use R to compute $P(X > 75)$ where $X \sim \mathcal{N}(\mu, \sigma^2)$. It may help to standardize by letting $Z := \frac{X - \mu}{\sigma} \sim \mathcal{N}(0, 1)$ and compute $P(Z > \frac{75-\mu}{\sigma})$ instead.