Probability of a (discrete) random variable

I'm having trouble understanding this question:

Quote:

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)^{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?}

Re: Probability of a (discrete) random variable

Re: Probability of a (discrete) random variable

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

Re: Probability of a (discrete) random variable

Quote:

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?

Re: Probability of a (discrete) random variable

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?

Re: Probability of a (discrete) random variable

Quote:

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?

Re: Probability of a (discrete) random variable

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