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) =
`(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 \{ {{(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.((1)/(8-0) )} \, dx`

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

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

E(X) =
`(1)/(8).((x^(2))/(2) )`

E(X) =
`(1)/(8) ((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).(1)/(8-0) } \, dx - ((8+0)/(2))^(2)`

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

V(X) =
`(1)/(8) ((x^(3))/(3) ) - 16`

V(X) =
`(1)/(8) ((8^(3))/(3) ) - 16`

V(X) =
`(64)/(3)` - 16

V(X) =
`(16)/(3)`

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