Thread: Probability of a (discrete) random variable

I'm having trouble understanding this question:

Suppose the random variable x = 1, 2, 3, 4, 5, 6.
In addition suppose
p(1)=p(2) and p(3)=p(4)=p(5)=p(6)
If the mean μ=4.00, then P(2 ≤ x ≤ 3) is:

So since all the values of x are laid out, x is a discrete random variable. Since there are only really two distinct probabilities, I'm guessing this would be a binomial distribution.

The binomial distribution formula is: p(x) = nCx * p^x * (1-p)<sup>n-x

and since the mean is given, i know the mean = n*p, but I don't know how I'd find either n or p since neither are given

any ideas?</sup>

Originally Posted by snypeshow

I'm having trouble understanding this question:
Suppose the random variable x = 1, 2, 3, 4, 5, 6.
In addition suppose
p(1)=p(2) and p(3)=p(4)=p(5)=p(6)
If the mean μ=4.00, then P(2 ≤ x ≤ 3) is:

This is not binomial.
You have two probabilities .
From the given and .
Solve for . The your answer is

I understand how you got 2a and 4b, but how did you get 3a and 18b?

Originally Posted by snypeshow

I understand how you got 2a and 4b, but how did you get 3a and 18b?

How does one find the mean(expected value) of a distribution?

Expected value is the sum of x multiplied by its probabilities

(1+2)a and (3+4+5+6)b

ok, so I understand that part, but why would you set it to equal 4.0 which is the standard deviation?

Originally Posted by snypeshow

Expected value is the sum of x multiplied by its probabilities

(1+2)a and (3+4+5+6)b

ok, so I understand that part, but why would you set it to equal 4.0 which is the standard deviation?

Oh, ok, I gotchya!! Thanks dude!!