# Thread: Need help with discrete probabilities

1. ## Need help with discrete probabilities

Hey guys, I need help with 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(μ - 2σ ≤ x ≤ μ + 2σ) is:
The correct answer is 1.000, but I'm having trouble figuring out how to do it

from the question, i can derive two formulas

(1) 2a + 4b = 1
(2) 3a + 18b = 4.00

Formula (2) was derived from the expected value rule

not sure how to continue on from here... any help?

2. ## Re: Need help with discrete probabilities

Solve the simultaneous equations. You now know the probability of each value. Work out the standard deviation from the formula
(SD)^2=sum of (px^2)-(mean)^2.

3. ## Re: Need help with discrete probabilities

What do you mean by this:

Originally Posted by biffboy
(SD)^2=sum of (px^2)-(mean)^2.
σ2 = [Σ(x-µ)2]/n

what is (px^2) supposed to be?

4. ## Re: Need help with discrete probabilities

Let p1 = prob. x=1, p2= prob.x=2 and so on. We need the sum of px^2
That is p1(1)^2+p2(2)^2..........p6(6)^2 Then (sd)^2 =(sum of px^2)-(mean)^2. Perhaps you haven't seen this formula before but it is the correct one to use here.

5. ## Re: Need help with discrete probabilities

0.0833*12 + 0.0833*22 + 0.20833*32 + 0.20833*42 + 0.20833*52 + 0.20833*62 = 18.333

is this the expected value formula you're using?

6. ## Re: Need help with discrete probabilities

Yes that's correct. So for the standard deviation subtract (mean)^2 (which is 16) and take square root. I get 1.528
In probability distributions of this kind learn the formulae
Mean = Sigma px and (SD)^2= sigma px^2-(mean)^2