Finite Math Examples

$\begin{array}{c} x & P (x) \\ 0 & 0.12 \\ 1 & 0.22 \\ 2 & 0.17 \\ 3 & 0.13 \\ 4 & 0.11 \\ 5 & 0.1 \\ 6 & 0.13 \\ 7 & 0.02 \end{array}$

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

$0.12$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.12$ is between $0$ and $1$ inclusive

Step 1.3

$0.22$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.22$ is between $0$ and $1$ inclusive

Step 1.4

$0.17$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.17$ is between $0$ and $1$ inclusive

Step 1.5

$0.13$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.13$ is between $0$ and $1$ inclusive

Step 1.6

$0.11$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.11$ is between $0$ and $1$ inclusive

Step 1.7

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

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

Step 1.8

$0.13$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.13$ is between $0$ and $1$ inclusive

Step 1.9

$0.02$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.02$ is between $0$ and $1$ inclusive

Step 1.10

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 1.11

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

$0.12 + 0.22 + 0.17 + 0.13 + 0.11 + 0.1 + 0.13 + 0.02$

Step 1.12

The sum of the probabilities for all the possible $x$ values is $0.12 + 0.22 + 0.17 + 0.13 + 0.11 + 0.1 + 0.13 + 0.02 = 1$ .

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Step 1.12.1

Add $0.12$ and $0.22$ .

$0.34 + 0.17 + 0.13 + 0.11 + 0.1 + 0.13 + 0.02$

Step 1.12.2

Add $0.34$ and $0.17$ .

$0.51 + 0.13 + 0.11 + 0.1 + 0.13 + 0.02$

Step 1.12.3

Add $0.51$ and $0.13$ .

$0.64 + 0.11 + 0.1 + 0.13 + 0.02$

Step 1.12.4

Add $0.64$ and $0.11$ .

$0.75 + 0.1 + 0.13 + 0.02$

Step 1.12.5

Add $0.75$ and $0.1$ .

$0.85 + 0.13 + 0.02$

Step 1.12.6

Add $0.85$ and $0.13$ .

$0.98 + 0.02$

Step 1.12.7

Add $0.98$ and $0.02$ .

$1$

Step 1.13

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: $0.12 + 0.22 + 0.17 + 0.13 + 0.11 + 0.1 + 0.13 + 0.02 = 1$

The table satisfies the two properties of a probability distribution:

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

Property 2: $0.12 + 0.22 + 0.17 + 0.13 + 0.11 + 0.1 + 0.13 + 0.02 = 1$

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.

$Expectation = 0 \cdot 0.12 + 1 \cdot 0.22 + 2 \cdot 0.17 + 3 \cdot 0.13 + 4 \cdot 0.11 + 5 \cdot 0.1 + 6 \cdot 0.13 + 7 \cdot 0.02$

Step 3

Simplify the expression.

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply $0$ by $0.12$ .

$Expectation = 0 + 1 \cdot 0.22 + 2 \cdot 0.17 + 3 \cdot 0.13 + 4 \cdot 0.11 + 5 \cdot 0.1 + 6 \cdot 0.13 + 7 \cdot 0.02$

Step 3.1.2

Multiply $0.22$ by $1$ .

$Expectation = 0 + 0.22 + 2 \cdot 0.17 + 3 \cdot 0.13 + 4 \cdot 0.11 + 5 \cdot 0.1 + 6 \cdot 0.13 + 7 \cdot 0.02$

Step 3.1.3

Multiply $2$ by $0.17$ .

$Expectation = 0 + 0.22 + 0.34 + 3 \cdot 0.13 + 4 \cdot 0.11 + 5 \cdot 0.1 + 6 \cdot 0.13 + 7 \cdot 0.02$

Step 3.1.4

Multiply $3$ by $0.13$ .

$Expectation = 0 + 0.22 + 0.34 + 0.39 + 4 \cdot 0.11 + 5 \cdot 0.1 + 6 \cdot 0.13 + 7 \cdot 0.02$

Step 3.1.5

Multiply $4$ by $0.11$ .

$Expectation = 0 + 0.22 + 0.34 + 0.39 + 0.44 + 5 \cdot 0.1 + 6 \cdot 0.13 + 7 \cdot 0.02$

Step 3.1.6

Multiply $5$ by $0.1$ .

$Expectation = 0 + 0.22 + 0.34 + 0.39 + 0.44 + 0.5 + 6 \cdot 0.13 + 7 \cdot 0.02$

Step 3.1.7

Multiply $6$ by $0.13$ .

$Expectation = 0 + 0.22 + 0.34 + 0.39 + 0.44 + 0.5 + 0.78 + 7 \cdot 0.02$

Step 3.1.8

Multiply $7$ by $0.02$ .

$Expectation = 0 + 0.22 + 0.34 + 0.39 + 0.44 + 0.5 + 0.78 + 0.14$

Step 3.2

Simplify by adding numbers.

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Step 3.2.1

Add $0$ and $0.22$ .

$Expectation = 0.22 + 0.34 + 0.39 + 0.44 + 0.5 + 0.78 + 0.14$

Step 3.2.2

Add $0.22$ and $0.34$ .

$Expectation = 0.56 + 0.39 + 0.44 + 0.5 + 0.78 + 0.14$

Step 3.2.3

Add $0.56$ and $0.39$ .

$Expectation = 0.95 + 0.44 + 0.5 + 0.78 + 0.14$

Step 3.2.4

Add $0.95$ and $0.44$ .

$Expectation = 1.39 + 0.5 + 0.78 + 0.14$

Step 3.2.5

Add $1.39$ and $0.5$ .

$Expectation = 1.89 + 0.78 + 0.14$

Step 3.2.6

Add $1.89$ and $0.78$ .

$Expectation = 2.67 + 0.14$

Step 3.2.7

Add $2.67$ and $0.14$ .

$Expectation = 2.81$