Statistics Examples

\begin{matrix} x & P (x) 9 & 0.4 11 & 0.4 13 & 0.1 15 & 0.1 \end{matrix}

$\begin{array}{c} x & P (x) \\ 9 & 0.4 \\ 11 & 0.4 \\ 13 & 0.1 \\ 15 & 0.1 \end{array}$

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

$x$ 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

0.4

$0.4$ 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 1.3

0.1

$0.1$ is between

0

$0$ and

1

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

0.1

$0.1$ 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.

0.4 + 0.4 + 0.1 + 0.1

$0.4 + 0.4 + 0.1 + 0.1$

Step 1.6

The sum of the probabilities for all the possible

x

$x$ values is

0.4 + 0.4 + 0.1 + 0.1 = 1

$0.4 + 0.4 + 0.1 + 0.1 = 1$ .

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Step 1.6.1

Add

0.4

$0.4$ and

0.4

$0.4$ .

0.8 + 0.1 + 0.1

$0.8 + 0.1 + 0.1$

Step 1.6.2

Add

0.8

$0.8$ and

0.1

$0.1$ .

0.9 + 0.1

$0.9 + 0.1$

Step 1.6.3

Add

0.9

$0.9$ and

0.1

$0.1$ .

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

$x$ values

Property 2:

$0.4 + 0.4 + 0.1 + 0.1 = 1$

The table satisfies the two properties of a probability distribution:

Property 1:

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

$x$ values

Property 2:

$0.4 + 0.4 + 0.1 + 0.1 = 1$

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.

$Expectation = 9 \cdot 0.4 + 11 \cdot 0.4 + 13 \cdot 0.1 + 15 \cdot 0.1$

Step 3

Simplify the expression.

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply

$9$ by

$0.4$ .

$Expectation = 3.6 + 11 \cdot 0.4 + 13 \cdot 0.1 + 15 \cdot 0.1$

Step 3.1.2

Multiply

$11$ by

$0.4$ .

$Expectation = 3.6 + 4.4 + 13 \cdot 0.1 + 15 \cdot 0.1$

Step 3.1.3

Multiply

$13$ by

$0.1$ .

$Expectation = 3.6 + 4.4 + 1.3 + 15 \cdot 0.1$

Step 3.1.4

Multiply

$15$ by

$0.1$ .

$Expectation = 3.6 + 4.4 + 1.3 + 1.5$

Step 3.2

Simplify by adding numbers.

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Step 3.2.1

Add

$3.6$ and

$4.4$ .

$Expectation = 8 + 1.3 + 1.5$

Step 3.2.2

Add

$8$ and

$1.3$ .

$Expectation = 9.3 + 1.5$

Step 3.2.3

Add

$9.3$ and

$1.5$ .

$Expectation = 10.8$

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