Solving Probability: P(S<0), P(S=0), P(S>0)

[Kisa30]

Homework Statement

let x be a random variable with
P(X=-1) = 1/4
P(X=0) = 1/4
P(X=1) = 1/2
Let S be the sum of 25 independant random variables each with the same distribution as X.
Calculate approximately
a) P(S<0)
b) P(S=0)
c) P(S>0)

Homework Equations

Quite possibly the normal approximation.

The Attempt at a Solution

I'm not sure how to start this at all. Please help.

The answers are
a) P(S<0) = 0.05
b) P(S=0) = 0.03
c) P(S>0) = 0.92

Thanks in advanced!

 

[chiro]

Short of actually drawing a tree diagram for all possibilities or using the convolution theorem to obtain the CDF, I'm not exactly sure how you would do it.

Using a tree diagram is pretty easy, but it is a little monotonous to do. With this you simply draw every possibility given independence using P(A and B) = P(A) x P(B) you can add up all the probabilities and you should get the answer.

The convolution theorem allows you to obtain the CDF when you are trying to find the sum of multiple random variables of any distribution (for example X,Y, and Z you can find Prob(X + Y + Z) <= x where x is the value in the domain).

I do know about normal approximations but I can't see how you would apply it. If you find a distribution that has the independence of the binomial, but allows more than two events per trial, use that and then use properties about sums of the distributions to get another distribution which you can then use to get the required probabilities.

 

[HallsofIvy]

It is true that if a random variable X has any distribution with finite mean, $\mu$, and standard deviation, $\sigma$, then the sum of n observations from that population will have, approximately (with accuracy better for larger n) the normal distribution with mean $n\mu$ and standard deviation $\sigma\sqrt{n}$.

Here, X has mean $\mu= (1/4)(-1)+ (1/4)(0)+ (1/2)(1)= 1/4$ and variance (square of standard deviation) $(1/4)(-1- 1/4)^2+ (1/4)(0- 1/4)^2+ (1/2)(1- 1/4)^2=$$(1/4)(25/16)+$$(1/4)(1/16)+ (1/2)(9/16)= 44/64= 11/16$ so the standard deviation is $\sqrt{11}/4$. The sum of 25 observations will be approximately normally distributed with mean 25/4 and standard deviation $(5/4)\sqrt{11}$.

Since X is integer valued, the sum of 25 observations will also be integer valued and you should use the "half integer correction". That is, P(X= 0) should be P(-.5< X< .5).