A cat delivers $n$ kittens. Each one of them is a male with a probability of $1\over 2$, independently. Let $1\le k\le n$ be an integer. Denote $B$ as the event in which at least $k$ of the kittens are males. Let $k={n\over 2}+\sqrt{n}$. What is true about $P(B)$ as $n\to \infty$?
- It converges to $a\le 0.05$
- It converges to $0.25\le a\le 0.5$
- It converges to $a>0.5$
- None
Now first of all, why am I advised to use the CLT instead of computing $({1\over 2})^{k}$? Second of all, I am a bit stuck when using the CLT. Let $X_n=x_1+...+x_n$ be the number of male kittens out of $n$ kittens. That would mean that $x_i$ is an indicator and I get: $E[X_n]={n\over 2}$ and $Var(X_n)={n\over 4}$. Now, am I looking for $P(k-1\le X_n\le k+1)$? Because if so, I get: $P({{\sqrt{n}-1}\over {1\over 2}\sqrt{n}}\le {X_n-{n\over 2}\over {1\over 2}\sqrt{n}}\le {{\sqrt{n}+1}\over {1\over 2}\sqrt{n}})=\Phi({{\sqrt{n}+1}\over {1\over 2}\sqrt{n}})-\Phi({{\sqrt{n}-1}\over {1\over 2}\sqrt{n}})$. I don't know If I can compute the limit that way. I am really confused. I could use directions in this.
Since $n \to \infty$, we may assume that $n\ge 30$ in order to use the CLT. Thus, we can use the mean and variance from our given binomial distribution to build a normal approximation. As you've already noted $E[X_n] = n/2$ and $\sigma (X_n)) = \frac{\sqrt{n}}{2}$. Our goal is to find $P(x_n\ge k)=P(\frac{x_n-n/2}{.5\sqrt{n}})=$ $$P( \frac{x_n-n/2}{.5\sqrt{n}} \ge \frac{n/2+\sqrt{n}-n/2}{.5\sqrt{n}})=P(z\ge2)=1-P(z\le2)=1-.9772=.0228$$ The only thing that worried me about this is that a limit was never taken, however it was necessary to have $n\ge 30$ so that we could use CLT.