Finite Math Examples

$\begin{array}{c} x & P (x) \\ 5 & 0.1 \\ 10 & 0.2 \\ 13 & 0.3 \\ 16 & 0.1 \\ 17 & 0.3 \end{array}$

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 $x$ , $0 \leq P (x) \leq 1$ .

2. $P (x_{0}) + P (x_{1}) + P (x_{2}) + \dots + P (x_{n}) = 1$ .

Step 1.2

$0.1$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.1$ is between $0$ and $1$ inclusive

Step 1.3

$0.2$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.2$ is between $0$ and $1$ inclusive

Step 1.4

$0.3$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.3$ 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

$0.3$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.3$ 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.

$0 \leq P (x) \leq 1$ for all x values

Step 1.8

Find the sum of the probabilities for all the possible $x$ values.

$0.1 + 0.2 + 0.3 + 0.1 + 0.3$

Step 1.9

The sum of the probabilities for all the possible $x$ values is $0.1 + 0.2 + 0.3 + 0.1 + 0.3 = 1$ .

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Step 1.9.1

Add $0.1$ and $0.2$ .

$0.3 + 0.3 + 0.1 + 0.3$

Step 1.9.2

Add $0.3$ and $0.3$ .

$0.6 + 0.1 + 0.3$

Step 1.9.3

Add $0.6$ and $0.1$ .

$0.7 + 0.3$

Step 1.9.4

Add $0.7$ and $0.3$ .

$1$

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: $0 \leq P (x) \leq 1$ for all $x$ values

Property 2: $0.1 + 0.2 + 0.3 + 0.1 + 0.3 = 1$

The table satisfies the two properties of a probability distribution:

Property 1: $0 \leq P (x) \leq 1$ for all $x$ values

Property 2: $0.1 + 0.2 + 0.3 + 0.1 + 0.3 = 1$

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.

$u = 5 \cdot 0.1 + 10 \cdot 0.2 + 13 \cdot 0.3 + 16 \cdot 0.1 + 17 \cdot 0.3$

Step 3

Simplify each term.

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Step 3.1

Multiply $5$ by $0.1$ .

$u = 0.5 + 10 \cdot 0.2 + 13 \cdot 0.3 + 16 \cdot 0.1 + 17 \cdot 0.3$

Step 3.2

Multiply $10$ by $0.2$ .

$u = 0.5 + 2 + 13 \cdot 0.3 + 16 \cdot 0.1 + 17 \cdot 0.3$

Step 3.3

Multiply $13$ by $0.3$ .

$u = 0.5 + 2 + 3.9 + 16 \cdot 0.1 + 17 \cdot 0.3$

Step 3.4

Multiply $16$ by $0.1$ .

$u = 0.5 + 2 + 3.9 + 1.6 + 17 \cdot 0.3$

Step 3.5

Multiply $17$ by $0.3$ .

$u = 0.5 + 2 + 3.9 + 1.6 + 5.1$

Step 4

Simplify by adding numbers.

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Step 4.1

Add $0.5$ and $2$ .

$u = 2.5 + 3.9 + 1.6 + 5.1$

Step 4.2

Add $2.5$ and $3.9$ .

$u = 6.4 + 1.6 + 5.1$

Step 4.3

Add $6.4$ and $1.6$ .

$u = 8 + 5.1$

Step 4.4

Add $8$ and $5.1$ .

$u = 13.1$

Step 5

The variance of a distribution is a measure of the dispersion and is equal to the square of the standard deviation.

$s^{2} = \sum_{}^{} {(x - u)}^{2} \cdot (P (x))$

Step 6

Fill in the known values.

${(5 - (13.1))}^{2} \cdot 0.1 + {(10 - (13.1))}^{2} \cdot 0.2 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

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 $13.1$ .

${(5 - 13.1)}^{2} \cdot 0.1 + {(10 - (13.1))}^{2} \cdot 0.2 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.2

Subtract $13.1$ from $5$ .

${(- 8.1)}^{2} \cdot 0.1 + {(10 - (13.1))}^{2} \cdot 0.2 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.3

Raise $- 8.1$ to the power of $2$ .

$65.61 \cdot 0.1 + {(10 - (13.1))}^{2} \cdot 0.2 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.4

Multiply $65.61$ by $0.1$ .

$6.561 + {(10 - (13.1))}^{2} \cdot 0.2 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.5

Multiply $- 1$ by $13.1$ .

$6.561 + {(10 - 13.1)}^{2} \cdot 0.2 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.6

Subtract $13.1$ from $10$ .

$6.561 + {(- 3.1)}^{2} \cdot 0.2 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.7

Raise $- 3.1$ to the power of $2$ .

$6.561 + 9.61 \cdot 0.2 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.8

Multiply $9.61$ by $0.2$ .

$6.561 + 1.922 + {(13 - (13.1))}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.9

Multiply $- 1$ by $13.1$ .

$6.561 + 1.922 + {(13 - 13.1)}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.10

Subtract $13.1$ from $13$ .

$6.561 + 1.922 + {(- 0.1)}^{2} \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.11

Raise $- 0.1$ to the power of $2$ .

$6.561 + 1.922 + 0.01 \cdot 0.3 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.12

Multiply $0.01$ by $0.3$ .

$6.561 + 1.922 + 0.003 + {(16 - (13.1))}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.13

Multiply $- 1$ by $13.1$ .

$6.561 + 1.922 + 0.003 + {(16 - 13.1)}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.14

Subtract $13.1$ from $16$ .

$6.561 + 1.922 + 0.003 + {2.9}^{2} \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.15

Raise $2.9$ to the power of $2$ .

$6.561 + 1.922 + 0.003 + 8.41 \cdot 0.1 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.16

Multiply $8.41$ by $0.1$ .

$6.561 + 1.922 + 0.003 + 0.841 + {(17 - (13.1))}^{2} \cdot 0.3$

Step 7.1.17

Multiply $- 1$ by $13.1$ .

$6.561 + 1.922 + 0.003 + 0.841 + {(17 - 13.1)}^{2} \cdot 0.3$

Step 7.1.18

Subtract $13.1$ from $17$ .

$6.561 + 1.922 + 0.003 + 0.841 + {3.9}^{2} \cdot 0.3$

Step 7.1.19

Raise $3.9$ to the power of $2$ .

$6.561 + 1.922 + 0.003 + 0.841 + 15.21 \cdot 0.3$

Step 7.1.20

Multiply $15.21$ by $0.3$ .

$6.561 + 1.922 + 0.003 + 0.841 + 4.563$

Step 7.2

Simplify by adding numbers.

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Step 7.2.1

Add $6.561$ and $1.922$ .

$8.483 + 0.003 + 0.841 + 4.563$

Step 7.2.2

Add $8.483$ and $0.003$ .

$8.486 + 0.841 + 4.563$

Step 7.2.3

Add $8.486$ and $0.841$ .

$9.327 + 4.563$

Step 7.2.4

Add $9.327$ and $4.563$ .

$13.89$