Finite Math Examples

xP(x)10.450.180.210.1140.2
Step 1
Prove that the given table satisfies the two properties needed for a probability distribution.
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Step 1.1
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 1.2
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 1.3
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 1.4
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 1.5
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 1.6
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 1.7
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 1.8
Find the sum of the probabilities for all the possible x values.
0.4+0.1+0.2+0.1+0.2
Step 1.9
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 1.9.1
Add 0.4 and 0.1.
0.5+0.2+0.1+0.2
Step 1.9.2
Add 0.5 and 0.2.
0.7+0.1+0.2
Step 1.9.3
Add 0.7 and 0.1.
0.8+0.2
Step 1.9.4
Add 0.8 and 0.2.
1
1
Step 1.10
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
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
Step 2
The expectation mean of a distribution is the value expected if trials of the distribution could continue indefinitely. This is equal to each value multiplied by its discrete probability.
u=10.4+50.1+80.2+10.1+140.2
Step 3
Simplify each term.
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Step 3.1
Multiply 0.4 by 1.
u=0.4+50.1+80.2+10.1+140.2
Step 3.2
Multiply 5 by 0.1.
u=0.4+0.5+80.2+10.1+140.2
Step 3.3
Multiply 8 by 0.2.
u=0.4+0.5+1.6+10.1+140.2
Step 3.4
Multiply 0.1 by 1.
u=0.4+0.5+1.6+0.1+140.2
Step 3.5
Multiply 14 by 0.2.
u=0.4+0.5+1.6+0.1+2.8
u=0.4+0.5+1.6+0.1+2.8
Step 4
Simplify by adding numbers.
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Step 4.1
Add 0.4 and 0.5.
u=0.9+1.6+0.1+2.8
Step 4.2
Add 0.9 and 1.6.
u=2.5+0.1+2.8
Step 4.3
Add 2.5 and 0.1.
u=2.6+2.8
Step 4.4
Add 2.6 and 2.8.
u=5.4
u=5.4
Step 5
The variance of a distribution is a measure of the dispersion and is equal to the square of the standard deviation.
s2=(x-u)2(P(x))
Step 6
Fill in the known values.
(1-(5.4))20.4+(5-(5.4))20.1+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7
Simplify the expression.
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Step 7.1
Simplify each term.
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Step 7.1.1
Multiply -1 by 5.4.
(1-5.4)20.4+(5-(5.4))20.1+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.2
Subtract 5.4 from 1.
(-4.4)20.4+(5-(5.4))20.1+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.3
Raise -4.4 to the power of 2.
19.360.4+(5-(5.4))20.1+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.4
Multiply 19.36 by 0.4.
7.744+(5-(5.4))20.1+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.5
Multiply -1 by 5.4.
7.744+(5-5.4)20.1+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.6
Subtract 5.4 from 5.
7.744+(-0.4)20.1+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.7
Raise -0.4 to the power of 2.
7.744+0.160.1+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.8
Multiply 0.16 by 0.1.
7.744+0.016+(8-(5.4))20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.9
Multiply -1 by 5.4.
7.744+0.016+(8-5.4)20.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.10
Subtract 5.4 from 8.
7.744+0.016+2.620.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.11
Raise 2.6 to the power of 2.
7.744+0.016+6.760.2+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.12
Multiply 6.76 by 0.2.
7.744+0.016+1.352+(1-(5.4))20.1+(14-(5.4))20.2
Step 7.1.13
Multiply -1 by 5.4.
7.744+0.016+1.352+(1-5.4)20.1+(14-(5.4))20.2
Step 7.1.14
Subtract 5.4 from 1.
7.744+0.016+1.352+(-4.4)20.1+(14-(5.4))20.2
Step 7.1.15
Raise -4.4 to the power of 2.
7.744+0.016+1.352+19.360.1+(14-(5.4))20.2
Step 7.1.16
Multiply 19.36 by 0.1.
7.744+0.016+1.352+1.936+(14-(5.4))20.2
Step 7.1.17
Multiply -1 by 5.4.
7.744+0.016+1.352+1.936+(14-5.4)20.2
Step 7.1.18
Subtract 5.4 from 14.
7.744+0.016+1.352+1.936+8.620.2
Step 7.1.19
Raise 8.6 to the power of 2.
7.744+0.016+1.352+1.936+73.960.2
Step 7.1.20
Multiply 73.96 by 0.2.
7.744+0.016+1.352+1.936+14.792
7.744+0.016+1.352+1.936+14.792
Step 7.2
Simplify by adding numbers.
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Step 7.2.1
Add 7.744 and 0.016.
7.76+1.352+1.936+14.792
Step 7.2.2
Add 7.76 and 1.352.
9.112+1.936+14.792
Step 7.2.3
Add 9.112 and 1.936.
11.048+14.792
Step 7.2.4
Add 11.048 and 14.792.
25.84
25.84
25.84
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