Thread: Probability of a (discrete) random variable

1. Probability of a (discrete) random variable

I'm having trouble understanding this question:

Suppose the random variable x = 1, 2, 3, 4, 5, 6.
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 * px * (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?

2. Re: Probability of a (discrete) random variable

Originally Posted by snypeshow
I'm having trouble understanding this question:
Suppose the random variable x = 1, 2, 3, 4, 5, 6.
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 $\displaystyle a~\&~b$.
From the given $\displaystyle 2a+4b=1$ and $\displaystyle 3a+18b=4$.
Solve for $\displaystyle a~\&~b$. The your answer is $\displaystyle a+b$

3. Re: Probability of a (discrete) random variable

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

4. Re: Probability of a (discrete) random variable

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?

5. 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?

6. Re: Probability of a (discrete) random variable

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?
$\displaystyle \text{The standard deviation }\ne\text{ the mean.}$

7. Re: Probability of a (discrete) random variable

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