Finite Math Examples

$\begin{array}{c} x & P (x) \\ 1 & - 13 \\ 2 & - 31 \\ 3 & - 59 \\ 4 & - 97 \\ 5 & - 146 \end{array}$

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

$- 13$ is not greater than or equal to $0$ , which doesn't meet the first property of the probability distribution.

$- 13$ is not greater than or equal to $0$

Step 1.3

$- 31$ is not greater than or equal to $0$ , which doesn't meet the first property of the probability distribution.

$- 31$ is not greater than or equal to $0$

Step 1.4

$- 59$ is not greater than or equal to $0$ , which doesn't meet the first property of the probability distribution.

$- 59$ is not greater than or equal to $0$

Step 1.5

$- 97$ is not greater than or equal to $0$ , which doesn't meet the first property of the probability distribution.

$- 97$ is not greater than or equal to $0$

Step 1.6

$- 146$ is not greater than or equal to $0$ , which doesn't meet the first property of the probability distribution.

$- 146$ is not greater than or equal to $0$

Step 1.7

The probability $P (x)$ does not fall between $0$ and $1$ inclusive for all $x$ values, which does not meet the first property of the probability distribution.

The table does not satisfy the two properties of a probability distribution

Step 2

The table does not satisfy the two properties of a probability distribution, which means that the variance can't be found using the given table.

Can't find the variance