Statistics Examples

Describe the Distribution's Two Properties
xP(x)00.2410.3420.2230.1340.0350.0160.03
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
0.24 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.24 is between 0 and 1 inclusive
Step 3
0.34 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.34 is between 0 and 1 inclusive
Step 4
0.22 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.22 is between 0 and 1 inclusive
Step 5
0.13 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.13 is between 0 and 1 inclusive
Step 6
0.03 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.03 is between 0 and 1 inclusive
Step 7
0.01 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.01 is between 0 and 1 inclusive
Step 8
0.03 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.03 is between 0 and 1 inclusive
Step 9
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 10
Find the sum of the probabilities for all the possible x values.
0.24+0.34+0.22+0.13+0.03+0.01+0.03
Step 11
The sum of the probabilities for all the possible x values is 0.24+0.34+0.22+0.13+0.03+0.01+0.03=1.
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Step 11.1
Add 0.24 and 0.34.
0.58+0.22+0.13+0.03+0.01+0.03
Step 11.2
Add 0.58 and 0.22.
0.8+0.13+0.03+0.01+0.03
Step 11.3
Add 0.8 and 0.13.
0.93+0.03+0.01+0.03
Step 11.4
Add 0.93 and 0.03.
0.96+0.01+0.03
Step 11.5
Add 0.96 and 0.01.
0.97+0.03
Step 11.6
Add 0.97 and 0.03.
1
1
Step 12
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: 0.24+0.34+0.22+0.13+0.03+0.01+0.03=1
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 [x2  12  π  xdx ] 
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