What is the probability that 83% or more of this sample...? This equation comes from Edgenuity's course of Statistics, and I am taking the course as a high school senior. I understand how to find the z-score and, in this case, not the standard deviation.

A cell phone provider has 85% of its customers rank their service as "satisfactory.” Nico takes a random sample of 75 customers from this cell phone provider. What is the probability that 83% or more of this sample ranks the provider’s service as "satisfactory”?


A. 0.314

B. 0.485

C. 0.562

D. 0.686

My guess with the observed value and mean value for the $z$-score equation would be to substitute $0.83$ for the former and 0.85 for the latter.
For the standard deviation, I believe I use the equation $\sqrt{np (1-p)}$, substituting $0.85$ for $p$ and $75$ for $n$. However, when I found the calculation to be $\approx 3.0923$, I thought it was higher than what I'm normally used to. And sure enough, when it was substituted into the $z$-score equation, the z-score was far too small. How do I find the correct answer? What methods were incorrect?
 A: Let's interpret an individual customer's statement of "satisfactory" or "unsatisfactory" as a random variable $X_i$ that is $1$ with probability $0.85$ and $0$ with probability $0.15$. Then, you are trying to get a close value for:
$$
\mathbb{P}(\frac{1}{75}\sum_{i=1}^{75} X_i \geq 0.83) = \mathbb{P}(\sum_{i=1}^{75} X_i \geq 62.25).
$$
The sum of the $X_i$ follows a binomial distribution $B(75,0.85)$ since each $X_i$ is a Bernoulli random variable. Since computing the exact probability of a binomial distribution may require a sum with many terms, it's easier to approximate with a normal distribution $\mathcal{N}(75\cdot 0.85,75\cdot 0.85\cdot 0.15)=\mathcal{N}(63.75,9.5625)$.
Now, the $z$-score of $62.25$ relative to this normal distribution is about $-0.1569$. The area to the right of this point in the standard normal distribution is about $0.56$. (Recall that the area to the right is what we are looking for since we are interested in satisfaction above the $0.83$ threshold).
A: The number $X$ satisfied
out of $n = 75$ randomly chosen customers
has $X \sim \mathsf{Binom}(n = 75, p = .85).$
You seek $P(X < 75(.83) = 1-P(X \le 75(.83)) = 0.6684.$
This answer uses an exact binomial computation in R, rather than a normal approximation.
1 - pbinom(75*.83, 75, .85)
[1] 0.6683943

The closest of the available answers is (D).
Depending on now non-integer $75(.83)$ is
interpreted in the normal approximation, that
answer may be slightly different that the
result from R above. [If there is a similar example in your text, follow the format and conventions of that.]
In the figure below the desired probability is the sum of the heights of the blue bars to the right of the vertical dotted line. Its normal approximation is the area under the density curve to the right of that line.

R code for figure.
hdr = "BINOM(75, .85) with Normal Approx."
plot(x, PDF, type="h", lwd=2, col="blue", main=hdr)
 mu = 75*.85;  sg = sqrt(mu*(1-.85))
 curve(dnorm(x,mu,sg), add=T, lwd=2, col="orange")
 abline(v = 75*.83, lwd=2, lty="dotted")
 abline(h = 0, col="green2")

