Question

Consider a distribution with the following probability density function
\[ f(x) = \begin{cases} 0.5, & 0 < x < 2 \\ 0.0, & \text{Otherwise} \end{cases} \]
Given that the mean of the above probability distribution is 1, the variance (rounded off to two decimal places) is ______.

Hint:

This is a uniform distribution on (0,2); use Var(X) = (b-a)^2/12 or E[X^2] - (E[X])^2.

Answer 1

Correct Answer: 0.32

Step 1: Identify the distribution.
The probability density function is
\[ f(x) = 0.5, \quad 0 < x < 2 \]
and zero elsewhere. This is a uniform distribution over the interval \((0, 2)\), since the density is a flat constant value across that range.

Step 2: Recall the formula for variance.
For any distribution, the variance is defined as
\[ \text{Var}(X) = E[X^2] - \left(E[X]\right)^2 \]
We are told \(E[X] = 1\) (the mean), so we only need to find \(E[X^2]\).

Step 3: Compute \(E[X^2]\).
By definition,
\[ E[X^2] = \int_0^2 x^2 f(x)\, dx = \int_0^2 x^2 (0.5)\, dx \]
Carry out the integration:
\[ E[X^2] = 0.5 \left[\frac{x^3}{3}\right]_0^2 = 0.5 \times \frac{8}{3} = \frac{4}{3} \]
Numerically,
\[ E[X^2] = \frac{4}{3} \approx 1.3333 \]

Step 4: Substitute into the variance formula.
\[ \text{Var}(X) = E[X^2] - (E[X])^2 = \frac{4}{3} - (1)^2 = \frac{4}{3} - 1 = \frac{1}{3} \]

Step 5: Round off the answer.
\[ \frac{1}{3} \approx 0.3333 \]
Rounded to two decimal places, this gives \(0.33\).

Final Answer:
The variance of the distribution is
\[ \boxed{0.33} \]