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?
Best Answer
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).