Question

A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the likelihood that people were absent 0, 1, 2, 3, 4, or 5 days last month.Number of days Absent | Probability 0 0.60 1 0.20 2 0.12 3 0.04 4 0.04 5 0.001. What is the mean number of days absent? What are the variance and standard deviation?2. For the following probability distribution: a) x|10|11|12|13|14b) P(x)|.1|.25|.3|.25|.13. Find the mean, variance and standard deviation for the following probability distribution.

Answer 1

(1)

`E(x) = 0.72`

`Var(x) = 1.1616`

`\sigma = 1.078`

(2)

`E(x) = 12`

`Var(x) = 1.3`

`\sigma = 1.14`

Explanation:

Solving (1):

`\begin{array}{ccccccc}{Days} & {0} & {1} & {2} & {3} & {4}& {5} \ \\ {Probability} & {0.60} & {0.20} & {0.12} & {0.04} & {0.04} & {0.00} \ \end{array}`

(a): Mean

This is calculated as:

`E(x) = \sum x * P(x)`

So, we have:

`E(x) = 0 * 0.60 + 1 * 0.20 + 2 * 0.12 + 3 * 0.04 + 4 * 0.04 + 5 * 0.00`

`E(x) = 0.72`

Solving (b): The variance

This is calculated as:

`Var(x) = E(x^2) - (E(x))^2`

Where:

`E(x) = 0.72`

and
`E(x^2) = \sum x^2 * P(x)`

So, we have:

`E(x^2) = 0^2 * 0.60 + 1^2 * 0.20 + 2^2 * 0.12 + 3^2 * 0.04 + 4^2 * 0.04 + 5^2 * 0.00`

`E(x^2) = 1.68`

So, we have:

`Var(x) = E(x^2) - (E(x))^2`

`Var(x) = 1.68 - 0.72^2`

`Var(x) = 1.1616`

Solving (c): The standard deviation.

This is calculated as:

`\sigma = √(Var(x))`

`\sigma = √(1.1616)`

`\sigma = 1.078`

Solving (2):

`\begin{array}{ccccccc}{x} & {10} & {11} & {12} & {13} & {14}& { } \ \\ {P(x)} & {0.1} & {0.25} & {0.3} & {0.25} & {0.1} & { } \ \end{array}`

(a): Mean

This is calculated as:

`E(x) = \sum x * P(x)`

So, we have:

`E(x) = 10 * 0.10 + 11 * 0.25 + 12 * 0.3 + 13 * 0.25 + 14 * 0.1`

`E(x) = 12`

Solving (b): The variance

This is calculated as:

`Var(x) = E(x^2) - (E(x))^2`

Where:

`E(x) = 12`

and
`E(x^2) = \sum x^2 * P(x)`

So, we have:

`E(x^2) = 10^2 * 0.10 + 11^2 * 0.25 + 12^2 * 0.3 + 13^2 * 0.25 + 14^2 * 0.1`

`E(x^2) = 145.3`

So, we have:

`Var(x) = E(x^2) - (E(x))^2`

`Var(x) = 145.3 - 12^2`

`Var(x) = 1.3`

Solving (c): The standard deviation.

This is calculated as:

`\sigma = √(Var(x))`

`\sigma = √(1.3)`

`\sigma = 1.14`