Answer:
[tex]\mu =2.16[/tex] --- Mean
[tex]\sigma^2 = 1.4944[/tex] -- Variance
[tex]\sigma = 1.222[/tex] --- Standard deviation
Step-by-step explanation:
Given
[tex]\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ P(x) & {0.12} & {0.2} & {0.2} & {0.36} & {0.12} \ \end{array}[/tex]
Solving (a): The population mean
This is calculated as:
[tex]\mu = \sum x * P(x)[/tex]
So, we have:
[tex]\mu =0*0.12 + 1 * 0.2 + 2 * 0.2 + 3 * 0.36 + 4 * 0.12[/tex]
[tex]\mu =2.16[/tex]
Solving (b): The population variance
First, calculate:
[tex]E(x^2)[/tex] using:
[tex]E(x^2) = \sum x^2 * P(x)[/tex]
So, we have:
[tex]E(x^2) = 0^2*0.12 + 1^2 * 0.2 + 2^2 * 0.2 + 3^2 * 0.36 + 4^2 * 0.12[/tex]
[tex]E(x^2) =6.160[/tex]
So, the population variance is:
[tex]\sigma^2 = E(x^2) - \mu^2[/tex]
[tex]\sigma^2 = 6.16 - 2.160^2[/tex]
[tex]\sigma^2 = 6.160 - 4.6656[/tex]
[tex]\sigma^2 = 1.4944[/tex]
Solving (c): The population standard deviation
This is calculated as:
[tex]\sigma = \sqrt{\sigma^2}[/tex]
[tex]\sigma = \sqrt{1.4944}[/tex]
[tex]\sigma = 1.222[/tex]
