# Probability distribution table

• Aug 23rd 2011, 08:19 PM
juliacoolness
Probability distribution table
A coin and a six sided die are thrown simultaneously. The random variable X is defined as follows: if the coin shows a head, then X is the score on the die; if the coin shows a tail, then X is twice the score on the die.
At the moment in class, we are doing probability distribution, and I need to put this information into a probability distribution table in order to complete the question. I know how to find the expected value and variance once the data is in the table, but I just need some help sorting it into it.
• Aug 23rd 2011, 09:40 PM
pickslides
Re: Probability distribution table
For heads the die results are: 1, 2, 3, ..., 6.

For tails the die results are: 2, 4, 6, ..., 12.

Now make a table where X is the result and find the probability

X p(X)

1 1/12
2 2/12
3 1/12
4 2/12
5 ...
6 ...
8 ...
10 ...
12 ...

Fill in the rest, what do you get?

After this we can find E(X) & Var(x)
• Aug 24th 2011, 12:34 AM
juliacoolness
Re: Probability distribution table
Thanks so much!
So I got
5 - 1/12
6 - 2/12
8 - 1/12
10 - 1/12
12 - 1/12

After that I found E(X)=5.25.
And I tried to get Var(X), but I couldn't get the answer that was in the textbook. I got 595/48. I thought I has the right answer, since i followed the right steps of taking away each of the numbers in the set by the mean number, then squaring all those numbers, adding them up and dividing it by 9, because there is 9 numbers in the set.
• Aug 24th 2011, 01:39 AM
pickslides
Re: Probability distribution table
Var(X)= E(X^2) - [E(X)]^2 = (1^2(1/12)+2^2(2/12)+...+12^2(1/12))- 5.25^2