Statistics Examples

Describe the Distribution's Two Properties
xP(x)10.450.180.210.1140.2
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
0.4 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.4 is between 0 and 1 inclusive
Step 3
0.1 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.1 is between 0 and 1 inclusive
Step 4
0.2 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.2 is between 0 and 1 inclusive
Step 5
0.1 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.1 is between 0 and 1 inclusive
Step 6
0.2 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.2 is between 0 and 1 inclusive
Step 7
For each x, the probability P(x) falls between 0 and 1 inclusive, which meets the first property of the probability distribution.
0P(x)1 for all x values
Step 8
Find the sum of the probabilities for all the possible x values.
0.4+0.1+0.2+0.1+0.2
Step 9
The sum of the probabilities for all the possible x values is 0.4+0.1+0.2+0.1+0.2=1.
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Step 9.1
Add 0.4 and 0.1.
0.5+0.2+0.1+0.2
Step 9.2
Add 0.5 and 0.2.
0.7+0.1+0.2
Step 9.3
Add 0.7 and 0.1.
0.8+0.2
Step 9.4
Add 0.8 and 0.2.
1
1
Step 10
For each x, the probability of P(x) falls between 0 and 1 inclusive. In addition, the sum of the probabilities for all the possible x equals 1, which means that the table satisfies the two properties of a probability distribution.
The table satisfies the two properties of a probability distribution:
Property 1: 0P(x)1 for all x values
Property 2: 0.4+0.1+0.2+0.1+0.2=1
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