Finite Math Examples

Describe the Distribution's Two Properties table[[x,P(x)],[30000,3],[60000,6],[90000,9],[120000,12],[150000,15]]
xP(x)3000036000069000091200001215000015
Step 1
A discrete random variable x takes a set of separate values (such as 0, 1, 2...). Its probability distribution assigns a probability P(x) to each possible value x. For each x, the probability P(x) falls between 0 and 1 inclusive and the sum of the probabilities for all the possible x values equals to 1.
1. For each x, 0P(x)1.
2. P(x0)+P(x1)+P(x2)++P(xn)=1.
Step 2
3 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
3 is not less than or equal to 1
Step 3
6 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
6 is not less than or equal to 1
Step 4
9 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
9 is not less than or equal to 1
Step 5
12 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
12 is not less than or equal to 1
Step 6
15 is not less than or equal to 1, which doesn't meet the first property of the probability distribution.
15 is not less than or equal to 1
Step 7
The probability P(x) does not fall between 0 and 1 inclusive for all x values, which does not meet the first property of the probability distribution.
The table does not satisfy the two properties of a probability distribution
 [x2  12  π  xdx ]