uniform distribution

[logistic_guy]

Given a continuous uniform distribution, show that

\(\displaystyle \bold{(a)} \ \mu = \frac{A + B}{2}\)

\(\displaystyle \bold{(b)} \ \sigma^2 = \frac{(B - A)^2}{12}\)

 

[khansaheb]

Given a continuous uniform distribution, show that

\(\displaystyle \bold{(a)} \ \mu = \frac{A + B}{2}\)

\(\displaystyle \bold{(b)} \ \sigma^2 = \frac{(B - A)^2}{12}\)

Please define

continuous uniform distribution

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[logistic_guy]

\(\displaystyle \bold{(a)}\)

A continuous uniform distribution has this probability density function:

\(\displaystyle f(x) = \frac{1}{B - A}\) in the interval \(\displaystyle A \leq x \leq B\)

\(\displaystyle \textcolor{green}{\bold{mean}} \rightarrow \mu\)

If a random variable \(\displaystyle X\) follows a continuous uniform distribution, then we have:

\(\displaystyle \mu = E(X) = \int_{-\infty}^{\infty} xf(x) \ dx\)


\(\displaystyle = \int_{A}^{B} \frac{x}{B - A} \ dx = \frac{1}{B - A}\frac{x^2}{2}\bigg|_A^B\)


\(\displaystyle = \frac{1}{B - A}\frac{B^2 - A^2}{2} = \frac{1}{B - A}\frac{(B - A)(B + A)}{2}\)

Then,

\(\displaystyle \mu = \textcolor{green}{\frac{B + A}{2}}\)