Finite Math Examples

\begin{matrix} x & P (x) 4 & 0.2 7 & 0.4 11 & 0.4 \end{matrix}

$\begin{array}{c} x & P (x) \\ 4 & 0.2 \\ 7 & 0.4 \\ 11 & 0.4 \end{array}$

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

$x$ takes a set of separate values (such as

0

$0$ ,

1

$1$ ,

2

$2$ ...). Its probability distribution assigns a probability

P (x)

$P (x)$ to each possible value

x

$x$ . For each

x

$x$ , the probability

P (x)

$P (x)$ falls between

0

$0$ and

1

$1$ inclusive and the sum of the probabilities for all the possible

x

$x$ values equals to

1

$1$ .

1. For each

x

$x$ ,

0 \leq P (x) \leq 1

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

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

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

Step 1.2

0.2

$0.2$ is between

0

$0$ and

1

$1$ inclusive, which meets the first property of the probability distribution.

0.2

$0.2$ is between

0

$0$ and

1

$1$ inclusive

Step 1.3

0.4

$0.4$ is between

0

$0$ and

1

$1$ inclusive, which meets the first property of the probability distribution.

0.4

$0.4$ is between

0

$0$ and

1

$1$ inclusive

Step 1.4

For each

x

$x$ , the probability

P (x)

$P (x)$ falls between

0

$0$ and

1

$1$ inclusive, which meets the first property of the probability distribution.

0 \leq P (x) \leq 1

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

Step 1.5

Find the sum of the probabilities for all the possible

x

$x$ values.

0.2 + 0.4 + 0.4

$0.2 + 0.4 + 0.4$

Step 1.6

The sum of the probabilities for all the possible

x

$x$ values is

0.2 + 0.4 + 0.4 = 1

$0.2 + 0.4 + 0.4 = 1$ .

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Step 1.6.1

Add

0.2

$0.2$ and

0.4

$0.4$ .

0.6 + 0.4

$0.6 + 0.4$

Step 1.6.2

Add

0.6

$0.6$ and

0.4

$0.4$ .

1

$1$

1

$1$

Step 1.7

For each

x

$x$ , the probability of

P (x)

$P (x)$ falls between

0

$0$ and

1

$1$ inclusive. In addition, the sum of the probabilities for all the possible

x

$x$ equals

1

$1$ , 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

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

x

$x$ values

Property 2:

0.2 + 0.4 + 0.4 = 1

$0.2 + 0.4 + 0.4 = 1$

The table satisfies the two properties of a probability distribution:

Property 1:

0 \leq P (x) \leq 1

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

x

$x$ values

Property 2:

0.2 + 0.4 + 0.4 = 1

$0.2 + 0.4 + 0.4 = 1$

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.

4 \cdot 0.2 + 7 \cdot 0.4 + 11 \cdot 0.4

$4 \cdot 0.2 + 7 \cdot 0.4 + 11 \cdot 0.4$

Step 3

Simplify each term.

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Step 3.1

Multiply

4

$4$ by

0.2

$0.2$ .

0.8 + 7 \cdot 0.4 + 11 \cdot 0.4

$0.8 + 7 \cdot 0.4 + 11 \cdot 0.4$

Step 3.2

Multiply

7

$7$ by

0.4

$0.4$ .

0.8 + 2.8 + 11 \cdot 0.4

$0.8 + 2.8 + 11 \cdot 0.4$

Step 3.3

Multiply

11

$11$ by

0.4

$0.4$ .

0.8 + 2.8 + 4.4

$0.8 + 2.8 + 4.4$

0.8 + 2.8 + 4.4

$0.8 + 2.8 + 4.4$

Step 4

Simplify by adding numbers.

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Step 4.1

Add

0.8

$0.8$ and

2.8

$2.8$ .

3.6 + 4.4

$3.6 + 4.4$

Step 4.2

Add

3.6

$3.6$ and

4.4

$4.4$ .

8

$8$

8

$8$

Step 5

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

s = \sqrt{\sum {(x - u)}^{2} \cdot (P (x))}

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

Step 6

Fill in the known values.

\sqrt{{(4 - (8))}^{2} \cdot 0.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}

$\sqrt{{(4 - (8))}^{2} \cdot 0.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7

Simplify the expression.

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Step 7.1

Multiply

- 1

$- 1$ by

8

$8$ .

\sqrt{{(4 - 8)}^{2} \cdot 0.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}

$\sqrt{{(4 - 8)}^{2} \cdot 0.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7.2

Subtract

8

$8$ from

4

$4$ .

\sqrt{{(- 4)}^{2} \cdot 0.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}

$\sqrt{{(- 4)}^{2} \cdot 0.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7.3

Raise

- 4

$- 4$ to the power of

2

$2$ .

\sqrt{16 \cdot 0.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}

$\sqrt{16 \cdot 0.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7.4

Multiply

16

$16$ by

0.2

$0.2$ .

\sqrt{3.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}

$\sqrt{3.2 + {(7 - (8))}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7.5

Multiply

- 1

$- 1$ by

8

$8$ .

\sqrt{3.2 + {(7 - 8)}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}

$\sqrt{3.2 + {(7 - 8)}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7.6

Subtract

8

$8$ from

7

$7$ .

\sqrt{3.2 + {(- 1)}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}

$\sqrt{3.2 + {(- 1)}^{2} \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7.7

Raise

$- 1$ to the power of

$2$ .

$\sqrt{3.2 + 1 \cdot 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7.8

Multiply

$0.4$ by

$1$ .

$\sqrt{3.2 + 0.4 + {(11 - (8))}^{2} \cdot 0.4}$

Step 7.9

Multiply

$- 1$ by

$8$ .

$\sqrt{3.2 + 0.4 + {(11 - 8)}^{2} \cdot 0.4}$

Step 7.10

Subtract

$8$ from

$11$ .

$\sqrt{3.2 + 0.4 + 3^{2} \cdot 0.4}$

Step 7.11

Raise

$3$ to the power of

$2$ .

$\sqrt{3.2 + 0.4 + 9 \cdot 0.4}$

Step 7.12

Multiply

$9$ by

$0.4$ .

$\sqrt{3.2 + 0.4 + 3.6}$

Step 7.13

Add

$3.2$ and

$0.4$ .

$\sqrt{3.6 + 3.6}$

Step 7.14

Add

$3.6$ and

$3.6$ .

$\sqrt{7.2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\sqrt{7.2}$

Decimal Form:

$2.68328157 \dots$

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