Question
Suppose $X$ is uniformly distributed on the interval $a$ to $b$. Show that$$E[X]=\frac{a+b}{2} \text { and } \sigma^2=\frac{(b-a)^2}{12} .$$
Step 1
For a random variable $X$ uniformly distributed over the interval $[a, b]$, the probability density function is given by: $$ f(x) = \frac{1}{b-a} \quad \text{for } a \leq x \leq b. $$ Show more…
Show all steps
Your feedback will help us improve your experience
Jacob Fry and 84 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Suppose X is a random variable uniformly distributed on the interval [1, b]. If E(X) = 6Var(X), find b?
Let $X$ be uniformly distributed on the interval $[1,2] .$ Find $E(1 / X) .$ Is $E(1 / X)=$ $1 / E(X) ?$
Given a continuous uniform distribution, show that (a) $\mu=\frac{A+B}{2}$ and (b) $\sigma^{2}=\frac{(B-A)^{2}}{12}$.
Some Continuous Probability Distributions
Applications of the Normal Distribution
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD