[Math] Normal distribution of multiple events

probability

I'm currently working on this problem and am having a bit of difficulty getting started. I found the normal distributions for both Mike and Jack with a score of over 225, but I'm unsure how to proceed with them. How should I be taking the normal distribution keeping in mind both of their scores will total to over 225?

Mike's bowling scores are normally distributed with mean 110 and standard deviation 13, while Jack's scores are normally distributed
with mean 135 and standard deviation 10. If Mike and Jack each bowl one game, find the probability that the total of their scores is
above 225 (assuming their scores are independent)

Any direction would be great! Thanks

Best Answer

Hint: If $X$ and $Y$ are independent, normally distributed, with means $\mu_X$, $\mu_Y$ and variances $\sigma_X^2$, $\sigma_Y^2$ respectively, then $X+Y$ has normal distribution, mean $\mu_X+\mu_Y$, and variance $\sigma_X^2+\sigma_Y^2$.

More generally, if $a$ and $b$ are constants, then $aX+bY$ is normally distributed, with mean $a\mu_X+b\mu_Y$, and variance $a^2\sigma_X^2+b^2\sigma_Y^2$.

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[Math] Probability – Calculate chance total score will be over or under a score line

OK, sorry for misunderstanding your point. So, your model predicts $PV=8$ with errors $\epsilon \sim N(-0.3,13)$. Or, this is equivalent to saying that you expect the actual value to be $AV \sim N(8.3,13)$. ($PV=AV+\epsilon$)

Now, to compute the probability that $X\leq 5.5$ (or $x>5.5$) under the hypothesis that your model is correct. This is

$$\mathbb{P}(X\leq 5.5|X \sim N(8.3,13)) \; .$$

I don't know what programming language or other computer program you use, but most math or statistics interpretative languages (Matlab, R, etc...) have the necessary commands inbuilt. So in R, you would use the command

pnorm(5.5,8.3,13)

which computes the cumulative normal distribution in point $5.5$ for mean and variance as given. For the probability bigger than $5.5$, you take

1-pnorm(5.5,8.3,13)

If you want the explicit formula for the computation, I can provide that too, but unless you want to make a lot of extra work implenting the functions yourself, I don't think that is really useful.

[Math] Normal Distribution and Conditional Probability

Yes, It should definitely work. I am assuming that you know how to calculate probabilities of normal random variables. If that is the case then calculating the probability should be trivial.
Because the problem now reduces to $$\frac{P(Y\gt 90)}{P(Y\gt 75)}$$ I'll continue from thereon.
To compute $P(Y\gt 90)$ you first compute $P(Y\lt 90)$. To do his you first figure out how many standard deviations away from the mean is $90$. This is very often called the z-score, which you are aware of. So, $90$ is $10$ points away from your mean ($80$), this implies that 90 is $\frac{10}{8}=0.8 \sigma$s away from your mean. If you look up the probability of this particular z- score on a table, you'll see that $P(Y\lt90)=\Phi(0.8)=0.788.$ So
$P(Y\gt90)=1-\Phi(0.8)=0.212.$

Similarly, $P(Y\lt75)=\Phi(-0.625)=.266$. So $P(Y\gt75)=1-\Phi(-0.625)=.266=0.734$

So $$\frac{P(Y\gt 90)}{P(Y\gt 75)}=\frac{0.212}{0.734}=0.289$$

I hope It helps

Best Answer

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[Math] Probability – Calculate chance total score will be over or under a score line

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