So this is a question about Normal Distribution (correct me if I am wrong), but I don't know how to solve it as I don't have variance and I'm not sure about mean. Help is highly appreciated.
Assume we have 2 candidates A and B applying for some position, and we estimate that 53% of the voters prefer the candidate B. We organize a poll, asking randomly chosen 100 voters about their preferences. What is the (approximate) probability that at least the half of that polled persons will favor the candidate A? Note: You can take Φ(0.601) ≈ 0.726.
Every vote can be seen as a bernoulli random variable, with $p=1-0.53=0.47$. Therefore the sum of the 100 votes for candidate $A$ is binominal distributed as $S\sim Bin(100, 0.47)$. with $\mu=47$ and $\sigma=\sqrt{100\cdot 0.47\cdot 0.53}$
Now it is asked for $P(S \geq 50)$. We can use the converse probability here. $P(S \geq 50)=1-P(S\leq 49)$
By applying $\text{central limit}$ theorem we can approximate the binomial distribution by the normal distribution.
$P(S \geq 50)=1-P(S\leq 49)\approx 1-\Phi \left( \frac{49-47}{\sqrt{100\cdot 0.47\cdot 0.53}} \right)=1-\Phi(0.401)$
Now we have the problem that the lecturer expect a different value, since it was calculated $P(S \geq 50)=1-P(S\leq 50)\approx 1-\Phi \left( \frac{50-47}{\sqrt{100\cdot 0.47\cdot 0.53}} \right)=1-\Phi(0.601)$
But this is wrong since $P(X\geq x)=1-P(X\leq x-1)$, if $X$ is binomial distributed. It´s up to you how you deal with this situation.