Find the standard deviation of this probability distribution.

Ответ:

mem8163

06.02.2022

Complete Question

Find the standard deviation of this probability distribution. Give your answer to at least 2 decimal places

x 1 2 3 4 5

P(x) 0 0.31 0.12 0.23 0.4

The value is $\sigma = 0.89$

Step-by-step explanation:

From the question we are told that

The probability distribution is

x 1 2 3 4 5

P(x) 0 0.31 0.12 0.23 0.4

x^2 1 4 9 16 25

Generally the mean is mathematically represented as

$E(X) = \sum [x * P(x)]$

=> E(X) = (1 * 0) + (2 * 0.31 ) + (3 * 0.12 ) + (4 * 0.23) + (5 * 0.4 )

=> E(X) = 3.9

Generally

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

=> E(X^2) = (1 * 0) + (4 * 0.31 ) + (9 * 0.12) + (16 * 0.23) + (25 * 0.4)

=> E(X^2) =16

Generally the variance of this probability distribution is

V(X) = E(X^2) - [E(X)]^2

V(X) = 16 - 3.9^2

V(X) = 0.79

Generally the standard deviation of this probability distribution is

$\sigma = \sqrt{V(x)}$

=> $\sigma = \sqrt{0.79}$

=> $\sigma = 0.889$

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