How to find the standard deviation of a random variable


Asked by admin @ in Mathematics viewed by 296 People


Consider the discrete random variable X given in the table below. Calculate the mean, variance, and standard deviation of X . Also, calculate the expected value of X . Round solution to three decimal places, if necessary. x 6 8 9 15 P ( x ) 0.64 0.14 0.14 0.08 μ = σ 2 = σ = What is the expected value of X ? E ( X )

Answered by admin @



Answer:

Expected value or mean = E(x) = \mu = 7.42

Variance = \sigma^2 = 6.283

Standard deviation = \sigma = 2.506

Step-by-step explanation:

We are given the following information:

x       |      P(x)

6       |     0.64

8       |     0.14

9       |     0.14

15     |     0.08

The expected value or mean is given by

E(x) = \mu = x \cdot P(x) \\\\E(x) = \mu = 6 \cdot 0.64 + 8 \cdot 0.14 + 9 \cdot 0.14 + 15 \cdot 0.08 \\\\E(x) = \mu = 7.42

The variance is given by

\sigma^2 = \sum (x - \mu)^2 \cdot p(x)

\sigma^2 = (6 - 7.42)^2 \cdot 0.64 + (8 - 7.42)^2 \cdot 0.14 + (9 - 7.42)^2 \cdot 0.14 + (15 - 7.42)^2 \cdot 0.08 \\\\\sigma^2 = 1.291 + 0.0471 + 0.349 + 4.596 \\\\\sigma^2 = 6.283

The standard deviation is given by

\sigma = \sqrt{\sum (x - \mu)^2 \cdot p(x)} \\\\\sigma = \sqrt{\sigma^2} \\\\\sigma = \sqrt{6.283} \\\\\sigma = 2.506

Therefore,

Expected value or mean = E(x) = \mu = 7.42

Variance = \sigma^2 = 6.283

Standard deviation = \sigma = 2.506


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