Using CLT to calculate probability question

Question

Suppose $53$ percent of the population prefer red socks to green. If $100$ random people are asked, what is the probability that most (at least $50$) will say prefer GREEN:

So I set
\begin{align*} N & = 49.5\\ n & = 100\\ p & = 0.47 \end{align*}

Then.. \begin{align*} P(N \geq 49.5) & = P\left(\frac{N - np}{\sqrt{npq}}\right)\\ & \geq \frac{49.5 - 47}{\sqrt{100*.47*.53}}\\ & = \varphi(0.50)\\ & = 0.3085 \end{align*} Except, that's not the right answer. What did I do wrong in my calculations?

Answer 1

The "correct answer" using the binomial distribution is about $0.3078$ and you can find it with R using

1 - pbinom(49, 100, 0.47)

Your normal approximation with continuity correction giving $0.3085$ looks close enough, though my calculation suggests it could be $0.3082$ if you do not round in the middle of the caclution.

If the answer book is offering about $0.24134$ or $0.24157$ then it is using a different definition of "most". If it offering about $0.27389$ then it is not using a continuity correction.

Answer 2

Using CLT, the normal approximation to the binomial distribution will have mean $np$ and variance $np(1-p)$, where $n=100$ and $p=0.47$ (probability of people who prefer green).

What you want to find is $P(N\geq 49.5)$ (as a continuity correction of $-0.5$ needs to be applied) for the normal distribution, which will be $$P(N\geq 49.5)=1-\Phi\left(\frac{49.5-47}{\sqrt{47(1-0.47)}}\right)=0.30822$$