Statistics Examples

$\begin{array}{c} x & P (x) \\ 6 & 0.1 \\ 9 & 0.2 \\ 13 & 0.3 \\ 16 & 0.4 \end{array}$

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 $x$ 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

$0.4$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.4$ is between $0$ and $1$ inclusive

Step 1.6

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.

$0.1 + 0.2 + 0.3 + 0.4$

Step 1.8

The sum of the probabilities for all the possible $x$ values is $0.1 + 0.2 + 0.3 + 0.4 = 1$ .

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Step 1.8.1

Add $0.1$ and $0.2$ .

$0.3 + 0.3 + 0.4$

Step 1.8.2

Add $0.3$ and $0.3$ .

$0.6 + 0.4$

Step 1.8.3

Add $0.6$ and $0.4$ .

$1$

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

The table satisfies the two properties of a probability distribution:

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

Property 2: $0.1 + 0.2 + 0.3 + 0.4 = 1$

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.

$u = 6 \cdot 0.1 + 9 \cdot 0.2 + 13 \cdot 0.3 + 16 \cdot 0.4$

Step 3

Simplify each term.

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Step 3.1

Multiply $6$ by $0.1$ .

$u = 0.6 + 9 \cdot 0.2 + 13 \cdot 0.3 + 16 \cdot 0.4$

Step 3.2

Multiply $9$ by $0.2$ .

$u = 0.6 + 1.8 + 13 \cdot 0.3 + 16 \cdot 0.4$

Step 3.3

Multiply $13$ by $0.3$ .

$u = 0.6 + 1.8 + 3.9 + 16 \cdot 0.4$

Step 3.4

Multiply $16$ by $0.4$ .

$u = 0.6 + 1.8 + 3.9 + 6.4$

Step 4

Simplify by adding numbers.

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Step 4.1

Add $0.6$ and $1.8$ .

$u = 2.4 + 3.9 + 6.4$

Step 4.2

Add $2.4$ and $3.9$ .

$u = 6.3 + 6.4$

Step 4.3

Add $6.3$ and $6.4$ .

$u = 12.7$

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.

${(6 - (12.7))}^{2} \cdot 0.1 + {(9 - (12.7))}^{2} \cdot 0.2 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

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 $12.7$ .

${(6 - 12.7)}^{2} \cdot 0.1 + {(9 - (12.7))}^{2} \cdot 0.2 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.2

Subtract $12.7$ from $6$ .

${(- 6.7)}^{2} \cdot 0.1 + {(9 - (12.7))}^{2} \cdot 0.2 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.3

Raise $- 6.7$ to the power of $2$ .

$44.89 \cdot 0.1 + {(9 - (12.7))}^{2} \cdot 0.2 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.4

Multiply $44.89$ by $0.1$ .

$4.489 + {(9 - (12.7))}^{2} \cdot 0.2 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.5

Multiply $- 1$ by $12.7$ .

$4.489 + {(9 - 12.7)}^{2} \cdot 0.2 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.6

Subtract $12.7$ from $9$ .

$4.489 + {(- 3.7)}^{2} \cdot 0.2 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.7

Raise $- 3.7$ to the power of $2$ .

$4.489 + 13.69 \cdot 0.2 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.8

Multiply $13.69$ by $0.2$ .

$4.489 + 2.738 + {(13 - (12.7))}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.9

Multiply $- 1$ by $12.7$ .

$4.489 + 2.738 + {(13 - 12.7)}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.10

Subtract $12.7$ from $13$ .

$4.489 + 2.738 + {0.3}^{2} \cdot 0.3 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.11

Multiply ${0.3}^{2}$ by $0.3$ by adding the exponents.

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Step 7.1.11.1

Multiply ${0.3}^{2}$ by $0.3$ .

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Step 7.1.11.1.1

Raise $0.3$ to the power of $1$ .

$4.489 + 2.738 + {0.3}^{2} \cdot {0.3}^{1} + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.11.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4.489 + 2.738 + {0.3}^{2 + 1} + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.11.2

Add $2$ and $1$ .

$4.489 + 2.738 + {0.3}^{3} + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.12

Raise $0.3$ to the power of $3$ .

$4.489 + 2.738 + 0.027 + {(16 - (12.7))}^{2} \cdot 0.4$

Step 7.1.13

Multiply $- 1$ by $12.7$ .

$4.489 + 2.738 + 0.027 + {(16 - 12.7)}^{2} \cdot 0.4$

Step 7.1.14

Subtract $12.7$ from $16$ .

$4.489 + 2.738 + 0.027 + {3.3}^{2} \cdot 0.4$

Step 7.1.15

Raise $3.3$ to the power of $2$ .

$4.489 + 2.738 + 0.027 + 10.89 \cdot 0.4$

Step 7.1.16

Multiply $10.89$ by $0.4$ .

$4.489 + 2.738 + 0.027 + 4.356$

Step 7.2

Simplify by adding numbers.

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Step 7.2.1

Add $4.489$ and $2.738$ .

$7.227 + 0.027 + 4.356$

Step 7.2.2

Add $7.227$ and $0.027$ .

$7.254 + 4.356$

Step 7.2.3

Add $7.254$ and $4.356$ .

$11.61$