Suppose that $n=10^{6}$ people are nearly equally divided between two candidates, $A$ and $B$. Assume each person votes independently with probability $p=1/2$ for each candidate.
Problem+Solution A) The mean number of votes for $A$ will be $np = 10^6 \times (1/2) = 500000$ and sigma (standard deviation) will be $\sqrt{np(1-p)} = 500$.
Problem B) Let $Z$ be a standard normal, and $P(Z>z) = 0.025$. Find $z$. Solution: using the Z-table we have $z = 1.96$.
Problem C) Among the voters, there is a group of $N$ voters who can be highly motivated to vote for candidate A. How big does $N$ have to be to make the chance the candidate wins to be at least $99.75\%$?
Tthis part is where I'm having trouble with. I figure we let $N = mean + stand \times Z$ , so if we find the z score for $P(z > 99.75) = 2.81$. Thus, $N = 500,000 + 500 \times 2.81$.
However, the on the answer key, the answer is $N = 500,000 + 500\times 1.96$.
How is this the case when when we want it to be at least $99.75\%$? Why are we using a $z$ score of $1.96$?
Also, I tried using the tutorial when trying my equations, but it is not working for me
Assume $X\sim Bin(10^6,0.5)$ votes for $A$. Approximate this with $Y=(X-500000)/500$ and $Y\approx N(0,1)=:Z$. Since $$P(Z\ge -2.81)=99.75\%,$$ we have approximately $$P(X\ge 500000 - 500 \times 2.81)=99.75\%.$$
So you have to bribe $500\times2.81$ non-voters to make this the case. If you're bribing existing voters, half of them will already be $A$-voters, but the other half will make $B$ lose votes, so you end up with the same solution.