# Approximating a sum of two binomial distributions

A club basketball team will play a 60-game season. Thirty-two of these games are against class A teams, and 28 are against class B teams. The outcomes of all the games are independent. The team will win each game against a class A opponent with probability .5 and it will win each game against a class B opponent with probability .65. Let X denote its total number of victories in the season.

(a) What is the relationship between $X_A$, $X_B$, and $X$? Is $X$ a binomial random variable?

(b) Approximate the probability that the team wins 45 or more games this season.

What I did: $$X=X_A+X_B \\ X_A \text{ is a binomial random variable with } n=32 \text{ and } p=0.5 \\ X_B \text{ is a binomial random variable with } n=28 \text{ and } p=0.65 \\ p_X(k) = \sum_{a=0}^k P(X_A=a)P(X_B=k-a)$$ Can I express $p_X(k)$ as a binomial distribution somehow? Also, I think it is possible to approximate $P(X\ge45)$ using a normal distribution by the Central Limit Theorem, but I'm not sure how to apply it.

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## 1 Answer

$X$ is not a binomial random variable. One way to see this is using the moment generating function.

You can approximate $X$ by a normal random variable having the same mean and variance.
You may want to use the continuity correction, if your instructor has mentioned that.

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I know the mean for $X$ is $32(.5)+28(.65) = 34.2$, but how do I calculate the variance? – woaini Nov 11 '12 at 7:22
Variance of the sum of independent random variables is the sum of the variances.  Do you know (or can you look up) the formula for variance of a binomial random variable? – Robert Israel Nov 11 '12 at 8:07
Okay, so I found that the variance is $(32)(.5)(.5) + (28)(.65)(.35)=14.37$. Does applying the continuity correction mean that $P(X\ge45)=P(X>44)$? – woaini Nov 11 '12 at 8:24
Yes.  The continuity correction would mean using $P(X > 44.5)$ (i.e. we want to include $X=45$ but not $X=44$). – Robert Israel Nov 12 '12 at 0:41
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