Question

Given the probability density function f(x) = 1/5 over the interval [4, 9], find the expected value, the

Answer 1

Answer:

Step-by-step explanation:

Assume that f(x) = 0 for x outside the interval [4,9]. We will use the following

$E[X^k] = \int_{4}^{9}x^k f(x) dx$

$Var(X) = E[X^2}- (E[X])^2$

Standard deviation = $\sqrt[]{Var(X)}$

Mean = $E[X]$

Then,

$E[X] = \int_{4}^{9}\frac{1}{5}dx = \frac{9^2-4^2}{2\cdot 5} = \frac{13}{2}$

$E[X^2] = \int_{4}^{9}\frac{x^2}{5}dx = \frac{9^3-4^3}{3\cdot 5} = \frac{133}{3}$

Then, $Var(x) = \frac{133}{3}-(\frac{13}{2})^2 = \frac{25}{12}$

Then the standard deviation is $\frac{5}{2\sqrt[]{3}}$