Finite Math Examples

$\begin{array}{c} x & P (x) \\ 3 & 0.4 \\ 7 & 0.3 \\ 9 & 0.2 \\ 10 & 0.1 \end{array}$

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

$x$ ,

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

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

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

$0.3$ is between

$0$ and

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

$0.3$ is between

$0$ and

$1$ inclusive

Step 1.4

$0.2$ 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

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 1.7

Find the sum of the probabilities for all the possible

$x$ values.

$0.4 + 0.3 + 0.2 + 0.1$

Step 1.8

The sum of the probabilities for all the possible

$x$ values is

$0.4 + 0.3 + 0.2 + 0.1 = 1$ .

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Step 1.8.1

Add

$0.4$ and

$0.3$ .

$0.7 + 0.2 + 0.1$

Step 1.8.2

Add

$0.7$ and

$0.2$ .

$0.9 + 0.1$

Step 1.8.3

Add

$0.9$ and

$0.1$ .

$1$

Step 1.9

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:

$0.4 + 0.3 + 0.2 + 0.1 = 1$

The table satisfies the two properties of a probability distribution:

Property 1:

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

$x$ values

Property 2:

$0.4 + 0.3 + 0.2 + 0.1 = 1$

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.

$3 \cdot 0.4 + 7 \cdot 0.3 + 9 \cdot 0.2 + 10 \cdot 0.1$

Step 3

Simplify each term.

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Step 3.1

Multiply

$3$ by

$0.4$ .

$1.2 + 7 \cdot 0.3 + 9 \cdot 0.2 + 10 \cdot 0.1$

Step 3.2

Multiply

$7$ by

$0.3$ .

$1.2 + 2.1 + 9 \cdot 0.2 + 10 \cdot 0.1$

Step 3.3

Multiply

$9$ by

$0.2$ .

$1.2 + 2.1 + 1.8 + 10 \cdot 0.1$

Step 3.4

Multiply

$10$ by

$0.1$ .

$1.2 + 2.1 + 1.8 + 1$

Step 4

Simplify by adding numbers.

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Step 4.1

Add

$1.2$ and

$2.1$ .

$3.3 + 1.8 + 1$

Step 4.2

Add

$3.3$ and

$1.8$ .

$5.1 + 1$

Step 4.3

Add

$5.1$ and

$1$ .

$6.1$

Step 5

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

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

Step 6

Fill in the known values.

$\sqrt{{(3 - (6.1))}^{2} \cdot 0.4 + {(7 - (6.1))}^{2} \cdot 0.3 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7

Simplify the expression.

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Step 7.1

Multiply

$- 1$ by

$6.1$ .

$\sqrt{{(3 - 6.1)}^{2} \cdot 0.4 + {(7 - (6.1))}^{2} \cdot 0.3 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.2

Subtract

$6.1$ from

$3$ .

$\sqrt{{(- 3.1)}^{2} \cdot 0.4 + {(7 - (6.1))}^{2} \cdot 0.3 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.3

Raise

$- 3.1$ to the power of

$2$ .

$\sqrt{9.61 \cdot 0.4 + {(7 - (6.1))}^{2} \cdot 0.3 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.4

Multiply

$9.61$ by

$0.4$ .

$\sqrt{3.844 + {(7 - (6.1))}^{2} \cdot 0.3 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.5

Multiply

$- 1$ by

$6.1$ .

$\sqrt{3.844 + {(7 - 6.1)}^{2} \cdot 0.3 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.6

Subtract

$6.1$ from

$7$ .

$\sqrt{3.844 + {0.9}^{2} \cdot 0.3 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.7

Raise

$0.9$ to the power of

$2$ .

$\sqrt{3.844 + 0.81 \cdot 0.3 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.8

Multiply

$0.81$ by

$0.3$ .

$\sqrt{3.844 + 0.243 + {(9 - (6.1))}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.9

Multiply

$- 1$ by

$6.1$ .

$\sqrt{3.844 + 0.243 + {(9 - 6.1)}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.10

Subtract

$6.1$ from

$9$ .

$\sqrt{3.844 + 0.243 + {2.9}^{2} \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.11

Raise

$2.9$ to the power of

$2$ .

$\sqrt{3.844 + 0.243 + 8.41 \cdot 0.2 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.12

Multiply

$8.41$ by

$0.2$ .

$\sqrt{3.844 + 0.243 + 1.682 + {(10 - (6.1))}^{2} \cdot 0.1}$

Step 7.13

Multiply

$- 1$ by

$6.1$ .

$\sqrt{3.844 + 0.243 + 1.682 + {(10 - 6.1)}^{2} \cdot 0.1}$

Step 7.14

Subtract

$6.1$ from

$10$ .

$\sqrt{3.844 + 0.243 + 1.682 + {3.9}^{2} \cdot 0.1}$

Step 7.15

Raise

$3.9$ to the power of

$2$ .

$\sqrt{3.844 + 0.243 + 1.682 + 15.21 \cdot 0.1}$

Step 7.16

Multiply

$15.21$ by

$0.1$ .

$\sqrt{3.844 + 0.243 + 1.682 + 1.521}$

Step 7.17

Add

$3.844$ and

$0.243$ .

$\sqrt{4.087 + 1.682 + 1.521}$

Step 7.18

Add

$4.087$ and

$1.682$ .

$\sqrt{5.769 + 1.521}$

Step 7.19

Add

$5.769$ and

$1.521$ .

$\sqrt{7.29}$

Step 7.20

Rewrite

$7.29$ as

${2.7}^{2}$ .

$\sqrt{{2.7}^{2}}$

Step 7.21

Pull terms out from under the radical, assuming positive real numbers.

$2.7$

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