Question from an old test:
The bakery produces doughnuts with weights $W_1 \text{g}, W_2 \text{g}, W_3 \text{g}, \ldots$, where $W_i$ for each doughnut is drawn from a distribution with a finite expected value $\mu$ and variance $\sigma^2$. Let $p_n$ denote the probability that the weight of $n$ doughnuts exceeds $n \mu+5 \sigma \sqrt{n}$.
What can be said about the convergence of the sequence $p_n$, if:
- (a) the weight of each doughnut is independent, and $W_i$ has a normal distribution (weight can be negative).
- (b) the weight of each doughnut is independent, the distribution of $W_i$ is the same for each doughnut, but not necessarily normal.
- (c) each doughnut has the same weight $W_i=W$ drawn from a normal distribution.
(possible answers: (A) it is a constant sequence, (B) converges to $0$, (C) converges to $1$, (D) converges to $\frac{1}{2}$, (E) converges to another value, (F) divergent, (G) not enough information)
My answers are:
(a) - B, because it means that on average the weight of each doughnut will exceed $5$ standard deviations, and the more drawings there are, the smaller is the probability of this happening.
(b) - B, because from the CLT I know that it's the same situation as if we were choosing weights from a normal distribution (regardless of what the true distribution is), so reasoning is the same as in (a).
(c) - D, because if $n$ tends to infinity, then the term $\sigma \sqrt{n}$ in $n \mu+5 \sigma \sqrt{n}$ will be negligible, and the probability that we exceed the expected value is $\frac{1}{2}$. However, I have doubts about this, because in the same way one could say that in (a) and (b) it is also $\frac{1}{2}$.
Are these correct answers? If not, where is the flaw in the reasoning?
Edit:
After another try I got that in (a) and (b) the probability is constant and equal to $1- \Phi(5)$. I derived this from transforming the formula
$$
\lim_{n \rightarrow \infty} P(\frac{\sum_{i=1}^{n}(X_{i}) - n\mu}{\sqrt{n}\sigma} \leq \alpha) = \Phi(\alpha)
$$
to obtain
$$
\lim_{n \rightarrow \infty} P(\sum_{i=1}^{n}X_{i} \leq \alpha \cdot \sqrt{n}\sigma + n\mu) = \Phi(\alpha)
$$
$$ \lim_{n \rightarrow \infty} P(\sum_{i=1}^{n}X_{i} > 5 \cdot \sqrt{n}\sigma + n\mu) = 1-\Phi(5). $$
You need to think about what distribution the total weight of $n$ doughnuts has.
In (a) it is $N(n\mu,n\sigma^2)$, and so $p_n=1-\Psi(5)$.
In (b) it is only approximately $N(n\mu,n\sigma^2)$, and so $p_n\to1-\Psi(5)$.
In (c) it is $N(n\mu,n^2\sigma^2)$, and so $p_n=1-\Psi(5/\sqrt n)\to1-\Psi(0)=1/2$.