Question

g Find the mean and the variance of the random variable X with probability function or density f(x) of a uniform distribution on [0, 8].

Answer 1

Answer: E(X) = 4

              V(X) = $\frac{16}{3}$

Step-by-step explanation: An uniform distribution is a random variable X restricted to a finite interval [a,b] and has a constant function f(x) over this interval, i.e., the function is of form:

f(x) = $\left \{ {{\frac{1}{b-a} } \atop {0}} \right.$  

The mean or expectation of an unifrom distribution is:

E(X) = $\int\limits^b_a {x.f(x)} \, dx$

For the density function in interval [0,8], expectation value is:

E(X) = $\int\limits^8_0 {x.(\frac{1}{8-0} )} \, dx$

E(X) = $\int\limits^8_0 {\frac{x}{8} } \, dx$

E(X) = $\frac{1}{8}. \int\limits^8_0 {x} \, dx$

E(X) = $\frac{1}{8}.(\frac{x^{2}}{2} )$

E(X) = $\frac{1}{8} (\frac{8^{2}}{2} )$

E(X) = 4

Variance of a probability distribution can be written as:

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

For uniform distribution in interval [0,8]:

V(X) = $\int\limits^b_a {x^{2}.\frac{1}{8-0} } \, dx - (\frac{8+0}{2})^{2}$

V(X) = $\frac{1}{8} \int\limits^8_0 {x^{2}} \, dx - 4^{2}$

V(X) = $\frac{1}{8} (\frac{x^{3}}{3} ) - 16$

V(X) = $\frac{1}{8} (\frac{8^{3}}{3} ) - 16$

V(X) = $\frac{64}{3}$ - 16

V(X) = $\frac{16}{3}$

The mean and variance are 4 and 16/3, respectively