Question

Given the following discrete uniform probability distribution, find the expected value and standard deviation of the random variable. Round your final answer to three decimal places, if necessary. Probability Distributionx 0 1 2 3 4 5 6 7 8 9 P(X

Answer 1

Answer:

$E(x) = 4.500$ --- Expected value

$SD(x) = 2.872$ --- Standard deviation

Step-by-step explanation:

Given

$\begin{array}{ccccccccccc}x & {0} & {1} & {2} & {3} & {4}& {5} & {6} & {7} & {8} & {9} \ \\ P(X=x) & {\frac{1}{10}} & {\frac{1}{10}} & {\frac{1}{10}} & {\frac{1}{10}} & {\frac{1}{10}}& {\frac{1}{10}} & {\frac{1}{10}} & {\frac{1}{10}} & {\frac{1}{10}} & {\frac{1}{10}} \ \end{array}$

Solving (a): Expected value

This is calculated using:

$E(x) = \sum\limits^{9}_{i=0} x_i * P(X = x_i)$

Since they all have the same probability, the formula becomes:

$E(x) = \frac{1}{10}\sum\limits^{9}_{i=0} x_i$

$E(x) = \frac{1}{10}(0+1+2+3+4+5+6+7+8+9)$

$E(x) = \frac{1}{10}*45$

$E(x) = \frac{45}{10}$

$E(x) = 4.500$

Solving (b): Standard Deviation

First, we calculate the variance using

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

In (a), we have:

$E(x) = 4.500$

$E(x^2)$ is calculated as:

$E(x^2) = \sum\limits^{9}_{i=0} x_i^2 * P(X = x_i)$

Since they all have the same probability, the formula becomes:

$E(x^2) = \frac{1}{10}\sum\limits^{9}_{i=0} x_i^2$

So, we have:

$E(x^2) = \frac{1}{10}(0^2+1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2)$

Using a calculator

$E(x^2) = \frac{1}{10}(285)$

$E(x^2) = 28.5$

So:

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

$Var(x) = 28.5 - 4.5^2$

$Var(x) = 28.5 - 20.25$

$Var(x) = 8.25$

The standard deviation is then calculated as:

$SD(x) = \sqrt{Var(x)}$

$SD(x) = \sqrt{8.25}$

$SD(x) = 2.872$ ---- approximated