I'm currently studying Normal Distribution and z Scores, however, I'm struggling a little bit I wonder if anyone could help me with a problem so I that I can learn to do the rest of them.
This was the problem that came up.
Question: In the survey of 1095 people, 761 people said they voted in a recent presidential election. Voting records show that 67% of eligible voters actually did vote.
A) Find the probability that among 1095 randomly selected voters at least 761 actually did vote.
B) What do the results from part A) suggest?
Now I started out by finding my Probability which is 67% then I found the q which is 33%
then I identified the "n" which is 1095 I believe, Then I tried finding the mean (np) which gave me 733.65 then I tried finding the standard deviation which is sqrt(npq) which I obtained: 15.325 or 15.33
so then I plugged this numbers into the equation (1095-733.65)/15.33 = 23.57
that's suppose to be the z-score correct? however it can't be?? I'm so confused could someone help me.
We assume this data comes from a binomial distribution with parameter $p.$ The mean is $np$ and the variance is $npq.$ The actual z-score must be computed using the proportions, not the raw counts. That is, you should have
$$ z = \frac{\frac{761}{1095} - 0.67 \cdot 1095}{\sqrt{0.67 \cdot 0.33 \cdot 1095}}, $$
with $x=\frac{761}{1095}, \mu=0.67\cdot 1095,$ and $\sigma = \sqrt{0.67 \cdot 0.33 \cdot 1095}$. Do you see how I've used each piece? Hope this helps.