Use the probability distribution for the random variable x to answer the question. x 0 1 2 3 4 p(x) 0.12 0.2 0.2 0.36 0.12 Calculate the population mean, variance, and standard deviation. (Round your standard deviation to three decimal places.)

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]

can anyone help me and explain

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