Use the Normal Approximation to calculate $P(240 \leq T \leq 460)$

Question

Andy and Novak are playing a tennis match. I think Novak is going to win, but my friend is not so sure. I offer to give my friend £10 if Andy wins, as long as my friend gives me £20 if Novak wins. Suppose that the probability that Novak wins is 0.8.

1. Show that the expected value of my winnings, W, is £14 and the standard deviation of W is £12. (I can do this part)

2. Suppose that we repeat this bet a total of 25 times, i.e, we bet on each of the 25 matches between Andy and Novak. Assuming that Novak's probability of winning remains 0.8 and my winnings from each bet are independent, calculate the mean and standard deviation of my total winnings $T$, where $T = \sum_{i=1}^{25} W_i$ and $W_i$ is my winnings from the $i^{th}$ bet.

3. Use a Normal Approximation to calculate $P(240 \leq T \leq 460)$. You may assume that you do not need to use a continuity correction. You are given the following R output to complete the calculation

$> pnorm(1.8333) = 0.966621$

My workings:

To find the mean and standard deviation of T, do you simply multiply the mean and the variance of W by 25 (the number of matches)? If so, the mean and standard deviation would equal £350 and £60.

I am unsure how to complete question 3. Since I assume you have to get a z value of 1.8333 but I can't make that work.

Any help would be greatly appreciated!

Answer 1

Your calculation of the mean and standard deviation are correct. Here are some hints for answering question $3$:

• If $\ x=460\ $, then the $\ z\ $ score corresponding to $\ x\ $ is $\ \frac{460-350}{60}=$$\,\frac{11}{6}\approx$$\,1.8333\ $. If $\ y=240\ $, then the $\ z\ $ score corresponding to $\ y\ $ is $\ \frac{240-350}{60}=$$\,\frac{{-}11}{6}\approx$$\,{-}1.8333\ $
• The R function pnorm is the cumulative distribution of a standard normal variate. That is, if $\ X\ $ is such a variate, then $$ \text{pnorm}(z)=P(X\le z)\ .$$ It satisfies the equation $\ \text{pnorm}(-z)=1-\text{pnorm}(z)\ $.

\begin{align}\hspace{0.7em}{\large\bullet}\ P(240\le T\le 460)&=P(T\le460)-P(T<240)\\&\approx\text{pnorm}(1.8333)-\text{pnorm}(-1.8333)\ .\end{align}

Answer 2

2. The reasoning here is that the mean of a sum of random variables is the sum of their means. Also, the variance of a sum of independent random variables is the sum of their variances.

3. Use the central limit theorem, which says for $W_i$ iid with mean $\mu$ and variance $\sigma^2$,

$$\frac{\frac{1}{n}\sum_{i=1}^nW_i-\mu}{\sigma/\sqrt n}\overset{d}{\rightarrow} N(0,1),$$

so that we have

$$P\left(a\leq \sum_{i=1}^nW_i\leq b\right)=P\left(\frac{a/n-\mu}{\sigma/\sqrt n}\leq \frac{\frac{1}{n}\sum_{i=1}^nW_i-\mu}{\sigma/\sqrt n}\leq \frac{b/n-\mu}{\sigma/\sqrt n}\right)\\ \approx P\left(\frac{a/n-\mu}{\sigma/\sqrt n}\leq Z\leq \frac{b/n-\mu}{\sigma/\sqrt n}\right)$$

for $Z\sim N(0,1).$