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You and your friend bowl every week. Your average score per game is normally distributed with an average value of $175$ and a standard deviation of $30.$ Your friend's score per game is normally distributed with an average value of $150$ and a standard deviation of $40.$ Assume that your points in a given game are independent. What is the probability that your friend in total will get more points than you over five weeks of play?

I can solve it only for a single case and I have got $p(z<0.5)=0.3085.$ How to define the probability that this even happens for $n=5?$ Is it relevant to calculate the probability for binomial distribution?

Air Mike
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Natasha
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  • I assume that $z$ denotes the ratio? I think you'll do better to consider the difference. Then you are just dealing with the sum of normal distributions. – lulu Sep 16 '20 at 11:38
  • I don't understand. – lulu Sep 16 '20 at 11:59
  • "Normally distributed" is a continuous distribution. You are adding normal distributions. There is no binomial distribution to be seen anywhere in this problem or in its solution. – David K Sep 16 '20 at 12:02
  • Sorry, lulu. It was my fault. I was on the right way but miscalculated standard deviations. I have received 0,1314. – Natasha Sep 16 '20 at 12:17

2 Answers2

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The question is "the probability that IN TOTAL your friend will have more points than you" over 5 weeks.

So you have to compare the two new Gaussian

$X\sim N(\mu;\sigma^2) = N(875;4500)$

$Y\sim N(\mu;\sigma^2) = N(750;8000)$

(Every gaussian X and Y is the sum of 5 iid Gaussian representing the weekly score)

and calculate $P(Y-X>0)$ as you know how getting

$P[(Y-X)>0]=1-\Phi(1.12)\approx 13.18\%$

tommik
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  • Thank you. But I have one more silly question to be sure. Shall I exclude P(x=0) when I calculate P(x=5) for binomial distribution? – Natasha Sep 16 '20 at 11:43
  • @NataliyaLutay : sorry I misunderstood the question...see my edited answer – tommik Sep 16 '20 at 11:52
  • Thank you, I have got approximately the same: 13,14%. I used the ordinary Z (or Φ) table for the normal distribution. – Natasha Sep 16 '20 at 12:32
  • @NataliyaLutay : ok. with the ordinary table I get exactly your result! The difference in result is due to the fact that the exact quantile is 1.118034... and not 1.12 – tommik Sep 16 '20 at 12:34
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The problem you have is that you are not clearly defining what you are trying to calculate.

Let $i \in \{1, 2, 3, 4, 5\}$ represent the week in which a game is played. We assume only one game is played per week. The score of the game in the $i^{\rm th}$ week is $X_i$ for you and $Y_i$ for your friend. Then we have $$X_i \sim \operatorname{Normal}(\mu_X = 175, \sigma_X = 30),$$ and $$Y_i \sim \operatorname{Normal}(\mu_Y = 150, \sigma_Y = 40).$$ Since we are asked to compare the total scores, over five weeks of play, it makes sense to define $$X = X_1 + X_2 + X_3 + X_4 + X_5, \\ Y = Y_1 + Y_2 + Y_3 + Y_4 + Y_5$$ as the sums of the scores for each respective player. Then the desired probability is $$\Pr[X < Y],$$ meaning your friend has a higher total score than you do.

Then, because the $X_i$ are independent and identically distributed normal random variables, $X$ is also a normal random variable whose mean equals $5\mu_X$, and whose standard deviation equals $\sqrt{5}\sigma_X$. And the corresponding situation is true for $Y$; i.e. $$Y \sim \operatorname{Normal}(5\mu_Y, \sqrt{5}\sigma_Y).$$ And since $X$ and $Y$ are independent, their difference is also normally distributed: $$Y - X \sim \operatorname{Normal}\left(\mu = 5\mu_Y - 5\mu_X, \sigma = \sqrt{5\sigma_Y^2 + 5\sigma_X^2}\right).$$ Note carefully that although the mean is the difference of means, the standard deviation of the variance is not. This is because for independent random variables, $$\operatorname{Var}[Y - X] = \operatorname{Var}[Y] + \operatorname{Var}[X].$$ Now that we know how the difference in total scores is distributed, the desired probability $\Pr[X < Y] = \Pr[Y - X > 0]$ is expressible as a statement about a normal distribution with a known mean and known standard deviation. We can then standardize and compare this to a standard normal probability table; e.g., $$\Pr[Y - X > 0] = \Pr\left[\frac{Y - X - \mu}{\sigma} > \frac{0 + 125}{50\sqrt{5}}\right] = \Pr[Z > 1.11803].$$

heropup
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