For this question, I have worked out that the Expected pay out per card is £0.90 and the variance is £24.99.
My approach: Let Sn be the total amount paid out from n cards, and we want the probability that Sn
Is it correct that Sn tends to a Normal distribution with mean 0.9n and variance 25n?
If this is correct, I'm not sure how to solve for n, since the equation contains an inequality that involves both Sn and n.
Thank you for any help given!
Let $X_n$ be the payout of the $n$-th ticket sold. All $X_n \sim X$ with $\mathbb{E}[X] = 0.9$ and $\operatorname{Var}(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = 24.99$.
Define $S_n := X_1 + X_2 + \cdots + X_n$. So, we want to determine the minimal positive integer $n$ such that
\begin{equation} \mathbb{P}(S_n < n) \ge 0.99. \end{equation}
This probability can be rewritten,
\begin{align} \mathbb{P}(S_n < n) = \mathbb{P}(S_n - 0.9n < 0.1n) = \mathbb{P}(\frac{S_n - 0.9n}{24.99 \sqrt{n}} < \frac{0.1n}{24.99 \sqrt{n}}). \end{align}
Set $Z_n = \frac{S_n - 0.9n}{24.99 \sqrt{n}}$. $Z_n$ approximately has a normal distribution. Now we wish to determine the minimal positive integer $n$ such that
\begin{equation} \mathbb{P}(Z_n < \frac{0.1n}{24.99 \sqrt{n}}) \ge 0.99. \end{equation}
This can be done using a software tool such as Matlab.