Finite Math Examples

\begin{array}{c} x & P (x) \\ 2 & 0.1 \\ 4 & 0.3 \\ 7 & 0.2 \\ 10 & 0.1 \\ 11 & 0.1 \\ 13 & 0.2 \end{array}

$\begin{array}{c} x & P (x) \\ 2 & 0.1 \\ 4 & 0.3 \\ 7 & 0.2 \\ 10 & 0.1 \\ 11 & 0.1 \\ 13 & 0.2 \end{array}$

Step 1

Prove that the given table satisfies the two properties needed for a probability distribution.

Tap for more steps...

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

0.1

$0.1$ 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 1.3

0.3

$0.3$ is between

0

$0$ and

1

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

0.3

$0.3$ is between

0

$0$ and

1

$1$ inclusive

Step 1.4

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 1.5

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 1.6

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 1.7

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 1.8

Find the sum of the probabilities for all the possible

x

$x$ values.

0.1 + 0.3 + 0.2 + 0.1 + 0.1 + 0.2

$0.1 + 0.3 + 0.2 + 0.1 + 0.1 + 0.2$

Step 1.9

The sum of the probabilities for all the possible

x

$x$ values is

0.1 + 0.3 + 0.2 + 0.1 + 0.1 + 0.2 = 1

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

Tap for more steps...

Step 1.9.1

Add

0.1

$0.1$ and

0.3

$0.3$ .

0.4 + 0.2 + 0.1 + 0.1 + 0.2

$0.4 + 0.2 + 0.1 + 0.1 + 0.2$

Step 1.9.2

Add

0.4

$0.4$ and

0.2

$0.2$ .

0.6 + 0.1 + 0.1 + 0.2

$0.6 + 0.1 + 0.1 + 0.2$

Step 1.9.3

Add

0.6

$0.6$ and

0.1

$0.1$ .

0.7 + 0.1 + 0.2

$0.7 + 0.1 + 0.2$

Step 1.9.4

Add

0.7

$0.7$ and

0.1

$0.1$ .

0.8 + 0.2

$0.8 + 0.2$

Step 1.9.5

Add

0.8

$0.8$ and

0.2

$0.2$ .

1

$1$

1

$1$

Step 1.10

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:

0.1 + 0.3 + 0.2 + 0.1 + 0.1 + 0.2 = 1

$0.1 + 0.3 + 0.2 + 0.1 + 0.1 + 0.2 = 1$

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:

0.1 + 0.3 + 0.2 + 0.1 + 0.1 + 0.2 = 1

$0.1 + 0.3 + 0.2 + 0.1 + 0.1 + 0.2 = 1$

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 = 2 \cdot 0.1 + 4 \cdot 0.3 + 7 \cdot 0.2 + 10 \cdot 0.1 + 11 \cdot 0.1 + 13 \cdot 0.2

$Expectation = 2 \cdot 0.1 + 4 \cdot 0.3 + 7 \cdot 0.2 + 10 \cdot 0.1 + 11 \cdot 0.1 + 13 \cdot 0.2$

Step 3

Simplify the expression.

Tap for more steps...

Step 3.1

Simplify each term.

Tap for more steps...

Step 3.1.1

Multiply

2

$2$ by

0.1

$0.1$ .

Expectation = 0.2 + 4 \cdot 0.3 + 7 \cdot 0.2 + 10 \cdot 0.1 + 11 \cdot 0.1 + 13 \cdot 0.2

$Expectation = 0.2 + 4 \cdot 0.3 + 7 \cdot 0.2 + 10 \cdot 0.1 + 11 \cdot 0.1 + 13 \cdot 0.2$

Step 3.1.2

Multiply

4

$4$ by

0.3

$0.3$ .

Expectation = 0.2 + 1.2 + 7 \cdot 0.2 + 10 \cdot 0.1 + 11 \cdot 0.1 + 13 \cdot 0.2

$Expectation = 0.2 + 1.2 + 7 \cdot 0.2 + 10 \cdot 0.1 + 11 \cdot 0.1 + 13 \cdot 0.2$

Step 3.1.3

Multiply

7

$7$ by

0.2

$0.2$ .

Expectation = 0.2 + 1.2 + 1.4 + 10 \cdot 0.1 + 11 \cdot 0.1 + 13 \cdot 0.2

$Expectation = 0.2 + 1.2 + 1.4 + 10 \cdot 0.1 + 11 \cdot 0.1 + 13 \cdot 0.2$

Step 3.1.4

Multiply

10

$10$ by

0.1

$0.1$ .

Expectation = 0.2 + 1.2 + 1.4 + 1 + 11 \cdot 0.1 + 13 \cdot 0.2

$Expectation = 0.2 + 1.2 + 1.4 + 1 + 11 \cdot 0.1 + 13 \cdot 0.2$

Step 3.1.5

Multiply

11

$11$ by

0.1

$0.1$ .

Expectation = 0.2 + 1.2 + 1.4 + 1 + 1.1 + 13 \cdot 0.2

$Expectation = 0.2 + 1.2 + 1.4 + 1 + 1.1 + 13 \cdot 0.2$

Step 3.1.6

Multiply

13

$13$ by

0.2

$0.2$ .

Expectation = 0.2 + 1.2 + 1.4 + 1 + 1.1 + 2.6

$Expectation = 0.2 + 1.2 + 1.4 + 1 + 1.1 + 2.6$

Expectation = 0.2 + 1.2 + 1.4 + 1 + 1.1 + 2.6

$Expectation = 0.2 + 1.2 + 1.4 + 1 + 1.1 + 2.6$

Step 3.2

Simplify by adding numbers.

Tap for more steps...

Step 3.2.1

Add

0.2

$0.2$ and

1.2

$1.2$ .

Expectation = 1.4 + 1.4 + 1 + 1.1 + 2.6

$Expectation = 1.4 + 1.4 + 1 + 1.1 + 2.6$

Step 3.2.2

Add

1.4

$1.4$ and

1.4

$1.4$ .

Expectation = 2.8 + 1 + 1.1 + 2.6

$Expectation = 2.8 + 1 + 1.1 + 2.6$

Step 3.2.3

Add

2.8

$2.8$ and

1

$1$ .

Expectation = 3.8 + 1.1 + 2.6

$Expectation = 3.8 + 1.1 + 2.6$

Step 3.2.4

Add

3.8

$3.8$ and

1.1

$1.1$ .

Expectation = 4.9 + 2.6

$Expectation = 4.9 + 2.6$

Step 3.2.5

Add

4.9

$4.9$ and

2.6

$2.6$ .

Expectation = 7.5

$Expectation = 7.5$

Expectation = 7.5

$Expectation = 7.5$

Expectation = 7.5

$Expectation = 7.5$

Enter YOUR Problem