# Handicapping Statistics

Lets say, that the Los Angeles Dodgers have an average runs scored of 5.25 (the sample mean). The standard deviation across this sample is 3.15. The Colorado Rockies have an average runs scored of 4.45 and a standard deviation of 3.65. Assuming normal distributions, how would we determine the probability of the Rockies scoring more runs than the Dodgers, and vise versa.

I know that I am really generalizing things here...there is more to it. This example illustrates what I am trying to do though....compare two normal distrubutions. I have taken statistics courses, but it's been a while...any help on this would be greatly appreciated.

EnumaElish
Homework Helper
You're looking for the probability P(scoreD < scoreR) or vice versa.

P(scoreD < scoreR) = P(0 < scoreR - scoreD)

which means you need to derive the distribution of the difference scoreR - scoreD.

Luckily, the difference of two normals is itself a normal distribution. See Properties. In your case, the difference is V = X - Y = scoreR - scoreD.

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Clarification

I understand how you can find a z score for either of these distributions, but I don't know where to measure the score.

If I want to find the probability of the Dodgers scoring 4 runs...I could find the z score for that. Wouldn't I have to know where these two distributions meet though, before you can just subtract the results of the z score.

Thanks for your reply...if you could clarify this for me that would be great!

EnumaElish
Homework Helper
For this type of problem you cannot find the z scores first, then take their difference.

The mean of "R - D" is (mean of R) - (mean of D). Say the difference is M = 4.45 - 5.25.

The variance of "R - D" is (var. of R) + (var of D). Say the sum is V = (3.65)2 + (3.15)2.

Now normalize z = (x - M)/sqrt(V). This is your z score for "R - D".

Say you want P(0 < R - D) = 1 - P(R - D < 0). That is x = 0. The z score you want is then z = (0 - M)/sqrt(V) = -M/sqrt(V). Which will give you the probability that the Rockies will "beat" the Dodgers.

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