What is the probability that 83% or more of this sample…?

Question

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 np (1 - p), substituting 0.85 for p and 75 for n. However, when I found the calculation to be 9.5625, 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?

Answer 1

It is greater than or equal, so it will be 1 - the probability. Also, sigma = sqrt(p(1-p)/n)

Thus, z = (.83-.85)/sqrt(.85(1-.85)/75) = -.4851, and 1 - Φ(-.4851) = Φ(.4851) = .6862