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lesya [120]
3 years ago
15

Given the following discrete uniform probability distribution, find the expected value and standard deviation of the random vari

able. Round your final answer to three decimal places, if necessary. Probability Distributionx 0 1 2 3 4 5 6 7 8 9 P(X
Mathematics
1 answer:
romanna [79]3 years ago
3 0

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

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