Question 29.11: A random variable, x, has p.d.f. f (x) given by f (x) = 1 0 ......
A random variable, x, has p.d.f. f (x) given by
f (x) = 1 0 ≤ x ≤ 1
Calculate the standard deviation of x.
Step-by-Step
The expected value, μ, is found:
\mu=\int_{0}^{1}x f(x)\,\mathrm{d}x=\left[{\frac{x^{2}}{2}}\right]_{0}^{1}={\frac{1}{2}}The variance can now be found:
variance = \sigma^{2}=\int_{0}^{1}\left(x-{\frac{1}{2}}\right)^{2}1\,\mathrm{d}x \\
=\int_{0}^{1}x^{2}-x+{\frac{1}{4}}\,\mathrm{d}x \\ =\left[{\frac{x^{3}}{3}}-{\frac{x^{2}}{2}}+{\frac{x}{4}}\right]_{0}^{1}={\frac{1}{3}}-{\frac{1}{2}}+{\frac{1}{4}}={\frac{1}{12}}Hence
\sigma=\sqrt{\frac{1}{12}}=0.29The standard deviation of x is 0.29.
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