Question

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 1

Answer:

$\mu =2.16$ --- Mean

$\sigma^2 = 1.4944$ -- Variance

$\sigma = 1.222$ --- Standard deviation

Step-by-step explanation:

Given

$\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ P(x) & {0.12} & {0.2} & {0.2} & {0.36} & {0.12} \ \end{array}$

Solving (a): The population mean

This is calculated as:

$\mu = \sum x * P(x)$

So, we have:

$\mu =0*0.12 + 1 * 0.2 + 2 * 0.2 + 3 * 0.36 + 4 * 0.12$

$\mu =2.16$

Solving (b): The population variance

First, calculate:

$E(x^2)$ using:

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2*0.12 + 1^2 * 0.2 + 2^2 * 0.2 + 3^2 * 0.36 + 4^2 * 0.12$

$E(x^2) =6.160$

So, the population variance is:

$\sigma^2 = E(x^2) - \mu^2$

$\sigma^2 = 6.16 - 2.160^2$

$\sigma^2 = 6.160 - 4.6656$

$\sigma^2 = 1.4944$

Solving (c): The population standard deviation

This is calculated as:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{1.4944}$

$\sigma = 1.222$