3. Let \( X \) be a r.v.; where \( X \sim \operatorname{Beta}(\alpha, \beta) \);and \( f(x)=30 x^{4}(1-x) ; 0 \leq x \leq 1 \). Find \( \mu \) and \( \sigma \) ?

To find the mean \( \mu \) and standard deviation \( \sigma \) of a random variable \( X \) that follows the Beta distribution with parameters \( \alpha \) and \( \beta \), we first need to identify the parameters from the provided probability density function \( f(x) = 30x^{4}(1-x) \). The Beta distribution \( \text{Beta}(\alpha, \beta) \) has the following properties for the mean and variance: 1. The mean \( \mu \) is given by: \[ \mu = \frac{\alpha}{\alpha + \beta} \] 2. The variance \( \sigma^2 \) is given by: \[ \sigma^2 = \frac{\alpha \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)} \] From the function \( f(x) = 30x^{4}(1-x) \), we can see that it is proportional to the Beta distribution. To express \( f(x) \) in the Beta function form, we note that it matches \( f(x) = \text{Beta}(\alpha, \beta) \) where \( \alpha = 5 \) and \( \beta = 2 \), since it can be represented as: \[ f(x) = C x^{\alpha-1} (1 - x)^{\beta-1} \] for \( C = \text{normalization constant} \). Taking \( \alpha = 5 \) and \( \beta = 2 \): 1. **Mean \( \mu \)**: \[ \mu = \frac{5}{5 + 2} = \frac{5}{7} \] 2. **Variance \( \sigma^2 \)**: \[ \sigma^2 = \frac{5 \cdot 2}{(5 + 2)^2 (5 + 2 + 1)} = \frac{10}{7^2 \cdot 8} = \frac{10}{49 \cdot 8} = \frac{10}{392} = \frac{5}{196} \] Thus, the standard deviation \( \sigma = \sqrt{\frac{5}{196}} = \frac{\sqrt{5}}{14} \). To summarize: \[ \mu = \frac{5}{7}, \quad \sigma = \frac{\sqrt{5}}{14}. \]