Question

If the probability density function of a random variable is given by $$ f(x) = \begin{cases} 12x^2(1 - x), & 0 \leq x \leq 1 \\ 0, & \text{elsewhere} \end{cases} $$ then the mean and variance are respectively:

• A. \( 0.6, 0.4 \)
• B. \( 0.4, 0.6 \)
• C. \( 0.2, 0.6 \)
• D. \( 0.6, 0.2 \)

Hint:

When calculating the mean and variance of a continuous random variable, use the definitions of mean \( \mu = \int x f(x) \, dx \) and variance \( \sigma^2 = \int x^2 f(x) \, dx - \mu^2 \) with the appropriate probability density function.

Answer 1

The Correct Option is B

The mean \( \mu \) of a random variable is given by:
\[ \mu = \int_0^1 x f(x) \, dx = \int_0^1 x \cdot 12x^2(1 - x) \, dx = \int_0^1 12x^3(1 - x) \, dx. \] Expanding the integrand: \[ \mu = 12 \int_0^1 (x^3 - x^4) \, dx = 12 \left[ \frac{x^4}{4} - \frac{x^5}{5} \right]_0^1 = 12 \left( \frac{1}{4} - \frac{1}{5} \right) = 12 \times \frac{1}{20} = 0.6. \] Now, the variance \( \sigma^2 \) is given by:
\[ \sigma^2 = \int_0^1 x^2 f(x) \, dx - \mu^2. \] First, compute the second moment: \[ \int_0^1 x^2 f(x) \, dx = \int_0^1 x^2 \cdot 12x^2(1 - x) \, dx = \int_0^1 12x^4(1 - x) \, dx. \] Expanding the integrand: \[ \int_0^1 12x^4(1 - x) \, dx = 12 \int_0^1 (x^4 - x^5) \, dx = 12 \left[ \frac{x^5}{5} - \frac{x^6}{6} \right]_0^1 = 12 \left( \frac{1}{5} - \frac{1}{6} \right) = 12 \times \frac{1}{30} = 0.4. \] Thus, the variance is: \[ \sigma^2 = 0.4 - (0.6)^2 = 0.4 - 0.36 = 0.04. \] Thus, the mean is \( 0.4 \) and the variance is \( 0.6 \).