Answer:
(1)
[tex]E(x) = 0.72[/tex]
[tex]Var(x) = 1.1616[/tex]
[tex]\sigma = 1.078[/tex]
(2)
[tex]E(x) = 12[/tex]
[tex]Var(x) = 1.3[/tex]
[tex]\sigma = 1.14[/tex]
Step-by-step explanation:
Solving (1):
[tex]\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}[/tex]
(a): Mean
This is calculated as:
[tex]E(x) = \sum x * P(x)[/tex]
So, we have:
[tex]E(x) = 0 * 0.60 + 1 * 0.20 + 2 * 0.12 + 3 * 0.04 + 4 * 0.04 + 5 * 0.00[/tex]
[tex]E(x) = 0.72[/tex]
Solving (b): The variance
This is calculated as:
[tex]Var(x) = E(x^2) - (E(x))^2[/tex]
Where:
[tex]E(x) = 0.72[/tex]
and [tex]E(x^2) = \sum x^2 * P(x)[/tex]
So, we have:
[tex]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[/tex]
[tex]E(x^2) = 1.68[/tex]
So, we have:
[tex]Var(x) = E(x^2) - (E(x))^2[/tex]
[tex]Var(x) = 1.68 - 0.72^2[/tex]
[tex]Var(x) = 1.1616[/tex]
Solving (c): The standard deviation.
This is calculated as:
[tex]\sigma = \sqrt{Var(x)}[/tex]
[tex]\sigma = \sqrt{1.1616}[/tex]
[tex]\sigma = 1.078[/tex]
Solving (2):
[tex]\begin{array}{ccccccc}{x} & {10} & {11} & {12} & {13} & {14}& { } \ \\ {P(x)} & {0.1} & {0.25} & {0.3} & {0.25} & {0.1} & { } \ \end{array}[/tex]
(a): Mean
This is calculated as:
[tex]E(x) = \sum x * P(x)[/tex]
So, we have:
[tex]E(x) = 10 * 0.10 + 11 * 0.25 + 12 * 0.3 + 13 * 0.25 + 14 * 0.1[/tex]
[tex]E(x) = 12[/tex]
Solving (b): The variance
This is calculated as:
[tex]Var(x) = E(x^2) - (E(x))^2[/tex]
Where:
[tex]E(x) = 12[/tex]
and [tex]E(x^2) = \sum x^2 * P(x)[/tex]
So, we have:
[tex]E(x^2) = 10^2 * 0.10 + 11^2 * 0.25 + 12^2 * 0.3 + 13^2 * 0.25 + 14^2 * 0.1[/tex]
[tex]E(x^2) = 145.3[/tex]
So, we have:
[tex]Var(x) = E(x^2) - (E(x))^2[/tex]
[tex]Var(x) = 145.3 - 12^2[/tex]
[tex]Var(x) = 1.3[/tex]
Solving (c): The standard deviation.
This is calculated as:
[tex]\sigma = \sqrt{Var(x)}[/tex]
[tex]\sigma = \sqrt{1.3}[/tex]
[tex]\sigma = 1.14[/tex]
