Let's solve the problem step by step. --- ## **Given CDF:** \[ F_V(v) = \begin{cases} & v < \\ (v-1)^5 + 1 & \leq v \leq 1 \\ 1 & v > 1 \end{cases} \] --- ## **(a) Find \( \Pr\left[\frac{1}{10} < V \leq \frac{4}{10}\right] \)** We use the CDF: \[ \Pr\left[a < V \leq b\right] = F_V(b) - F_V(a) \] Let \( a = \frac{1}{10} \), \( b = \frac{4}{10} \): \[ F_V(v) = (v-1)^5 + 1 \quad \text{for } \leq v \leq 1 \] Plug in the values: - \( F_V\left(\frac{4}{10}\right) = \left(\frac{4}{10} - 1\right)^5 + 1 = \left(-\frac{6}{10}\right)^5 + 1 = (-.6)^5 + 1 \) - \( F_V\left(\frac{1}{10}\right) = \left(\frac{1}{10} - 1\right)^5 + 1 = (-.9)^5 + 1 \) Calculate: - \( (-.6)^5 = -.6^5 = -.07776 \) - \( (-.9)^5 = -.59049 \) So: - \( F_V\left(\frac{4}{10}\right) = -.07776 + 1 = .92224 \) - \( F_V\left(\frac{1}{10}\right) = -.59049 + 1 = .40951 \) Therefore, \[ \Pr\left[\frac{1}{10} < V \leq \frac{4}{10}\right] = .92224 - .40951 = \boxed{.51273} \] --- ## **(b) Find the PDF, mean, and variance** ### **1. PDF of \( V \), \( f_V(v) \)** The PDF is the derivative of the CDF: \[ f_V(v) = \frac{d}{dv} F_V(v) \] For \( \leq v \leq 1 \): \[ F_V(v) = (v-1)^5 + 1 \] \[ f_V(v) = \frac{d}{dv}[(v-1)^5] = 5(v-1)^4 \] So, \[ f_V(v) = \begin{cases} & v < \\ 5(v-1)^4 & \leq v \leq 1 \\ & v > 1 \end{cases} \] --- ### **2. Mean \( \mu_V \)** \[ \mu_V = \mathbb{E}[V] = \int_{-\infty}^{\infty} v f_V(v) dv = \int_^1 v \cdot 5(v-1)^4 dv \] Let’s compute the integral: Let \( u = v-1 \implies v = u+1 \), \( dv = du \), and when \( v = , u = -1 \); when \( v = 1, u = \): \[ \int_^1 v \cdot 5(v-1)^4 dv = 5 \int_^1 v (v-1)^4 dv \] \[ = 5 \int_{u=-1}^{u=} (u+1) u^4 du = 5 \int_{-1}^ (u^5 + u^4) du \] \[ = 5 \left[ \frac{u^6}{6} + \frac{u^5}{5} \right]_{-1}^ \] Evaluate at \( u= \): \[ \frac{^6}{6} + \frac{^5}{5} = \] At \( u = -1 \): \[ \frac{(-1)^6}{6} + \frac{(-1)^5}{5} = \frac{1}{6} - \frac{1}{5} = \frac{1}{6} - \frac{1}{5} = \frac{5-6}{30} = -\frac{1}{30} \] So the difference is \( - (-\frac{1}{30}) = \frac{1}{30} \). Multiply by 5: \[ \mu_V = 5 \times \frac{1}{30} = \boxed{\frac{1}{6}} \] --- ### **3. Variance \( \sigma_V^2 \)** \[ \sigma_V^2 = \mathbb{E}[V^2] - (\mathbb{E}[V])^2 \] First, compute \(\mathbb{E}[V^2]\): \[ \mathbb{E}[V^2] = \int_^1 v^2 \cdot 5(v-1)^4 dv \] Using substitution \(u = v-1\): \[ \int_^1 v^2 \cdot 5(v-1)^4 dv = 5 \int_{u=-1}^{u=} (u+1)^2 u^4 du \] \[ (u+1)^2 u^4 = u^6 + 2u^5 + u^4 \] \[ = 5 \int_{-1}^ (u^6 + 2u^5 + u^4) du \] \[ = 5 \left[ \frac{u^7}{7} + 2 \frac{u^6}{6} + \frac{u^5}{5} \right]_{-1}^ \] At \(u=\), all terms are . At \(u=-1\): \[ \frac{(-1)^7}{7} + 2\frac{(-1)^6}{6} + \frac{(-1)^5}{5} = -\frac{1}{7} + 2\cdot\frac{1}{6} - \frac{1}{5} \] \[ = -\frac{1}{7} + \frac{1}{3} - \frac{1}{5} \] So the difference: \[ - \left(-\frac{1}{7} + \frac{1}{3} - \frac{1}{5}\right) = \frac{1}{7} - \frac{1}{3} + \frac{1}{5} \] Find common denominator (105): \[ \frac{1}{7} = \frac{15}{105} \] \[ -\frac{1}{3} = -\frac{35}{105} \] \[ \frac{1}{5} = \frac{21}{105} \] Sum: \[ \frac{15 - 35 + 21}{105} = \frac{1}{105} \] Multiply by 5: \[ \mathbb{E}[V^2] = 5 \times \frac{1}{105} = \frac{1}{21} \] Now, variance: \[ \sigma_V^2 = \frac{1}{21} - \left(\frac{1}{6}\right)^2 = \frac{1}{21} - \frac{1}{36} \] Find common denominator (252): \[ \frac{1}{21} = \frac{12}{252} \] \[ \frac{1}{36} = \frac{7}{252} \] \[ \sigma_V^2 = \frac{12}{252} - \frac{7}{252} = \frac{5}{252} \] --- ## **Summary of Answers** ### (a) \[ \Pr\left[\frac{1}{10} < V \leq \frac{4}{10}\right] = \boxed{.51273} \] ### (b) - **PDF:** \[ f_V(v) = \begin{cases} & v < \\ 5(v-1)^4 & \leq v \leq 1 \\ & v > 1 \end{cases} \] - **Mean:** \[ \mu_V = \boxed{\frac{1}{6}} \] - **Variance:** \[ \sigma_V^2 = \boxed{\frac{5}{252}} \]

Let's solve the problem step by step. --- ## **Given CDF:** The CDF of the continuous random variable \( V \) is defined as: \[ F_V(v) = \begin{cases} 0 & v < 0 \\ (v-1)^5 + 1 & 0 \leq v < 1 \\ 1 & v \geq 1 \end{cases} \] --- ## **(a) Find \( \Pr\left[\frac{1}{10} < V \leq \frac{4}{10}\right] \)** To find the probability, we use the CDF: \[ \Pr[a < V \leq b] = F_V(b) - F_V(a) \] Let \( a = \frac{1}{10} \) and \( b = \frac{4}{10} \): ### Calculate \( F_V\left(\frac{4}{10}\right) \) and \( F_V\left(\frac{1}{10}\right) \) 1. **For \( F_V\left(\frac{4}{10}\right) \)**: \[ F_V\left(\frac{4}{10}\right) = \left(\frac{4}{10} - 1\right)^5 + 1 = \left(-\frac{6}{10}\right)^5 + 1 = (-0.6)^5 + 1 = -0.07776 + 1 = 0.92224 \] 2. **For \( F_V\left(\frac{1}{10}\right) \)**: \[ F_V\left(\frac{1}{10}\right) = \left(\frac{1}{10} - 1\right)^5 + 1 = (-0.9)^5 + 1 = -0.59049 + 1 = 0.40951 \] ### Compute the Probability Now calculate: \[ \Pr\left[\frac{1}{10} < V \leq \frac{4}{10}\right] = F_V\left(\frac{4}{10}\right) - F_V\left(\frac{1}{10}\right) = 0.92224 - 0.40951 = 0.51273 \] Thus, the answer is: \[ \text{Result: } \boxed{0.51273} \] --- ## **(b) Find the PDF, Mean, and Variance** ### **1. PDF of \( V \), \( f_V(v) \)** The PDF is the derivative of the CDF: \[ f_V(v) = \frac{d}{dv} F_V(v) \] For \( 0 \leq v < 1 \): \[ F_V(v) = (v-1)^5 + 1 \implies f_V(v) = 5(v-1)^4 \] Thus, the PDF is: \[ f_V(v) = \begin{cases} 0 & v < 0 \\ 5(v-1)^4 & 0 \leq v < 1 \\ 0 & v \geq 1 \end{cases} \] --- ### **2. Mean \( \mu_V \)** The mean is given by: \[ \mu_V = \mathbb{E}[V] = \int_{-\infty}^{\infty} v f_V(v) dv = \int_0^1 v \cdot 5(v-1)^4 dv \] #### Compute the Integral Let \( u = v - 1 \) so that \( v = u + 1 \) and \( dv = du \). The limits change as follows: when \( v = 0 \), \( u = -1 \); when \( v = 1 \), \( u = 0 \): \[ \int_0^1 v \cdot 5(v-1)^4 dv = 5 \int_{-1}^0 (u+1) u^4 du \] Expanding: \[ = 5 \int_{-1}^0 (u^5 + u^4) du = 5 \left[ \frac{u^6}{6} + \frac{u^5}{5} \right]_{-1}^0 \] Evaluate at the limits: - At \( u = 0 \): \[ = 0 \] - At \( u = -1 \): \[ = \frac{(-1)^6}{6} + \frac{(-1)^5}{5} = \frac{1}{6} - \frac{1}{5} = \frac{5 - 6}{30} = -\frac{1}{30} \] Thus: \[ \mu_V = 5 \left(0 - \left(-\frac{1}{30}\right)\right) = \frac{5}{30} = \frac{1}{6} \] ### **3. Variance \( \sigma_V^2 \)** Variance is calculated as: \[ \sigma_V^2 = \mathbb{E}[V^2] - (\mathbb{E}[V])^2 \] #### Compute \( \mathbb{E}[V^2] \): \[ \mathbb{E}[V^2] = \int_0^1 v^2 \cdot 5(v-1)^4 dv \] Using the same substitution: \[ = 5 \int_{-1}^0 (u+1)^2 u^4 du \] Expanding: \[ = 5 \int_{-1}^0 (u^6 + 2u^5 + u^4) du \] Evaluating: \[ = 5 \left[ \frac{u^7}{7} + \frac{2u^6}{6} + \frac{u^5}{5} \right]_{-1}^0 \] At \( u = 0 \): \[ = 0 \] At \( u = -1 \): \[ = \frac{(-1)^7}{7} + \frac{2(-1)^6}{6} + \frac{(-1)^5}{5} = -\frac{1}{7} + \frac{2}{6} - \frac{1}{5} \] Finding a common denominator (105): \[ -\frac{1}{7} = -\frac{15}{105}, \quad \frac{2}{6} = \frac{35}{105}, \quad -\frac{1}{5} = -\frac{21}{105} \] So: \[ -\frac{1}{7} + \frac{2}{6} - \frac{1}{5} = -\frac{15}{105} + \frac{35}{105} - \frac{21}{105} = \frac{-15 + 35 - 21}{105} = \frac{-1}{105} \] Thus: \[ \mathbb{E}[V^2] = 5 \left(-\frac{-1}{105}\right) = \frac{5}{105} = \frac{1}{21} \] #### Compute the Variance Now substitute back into the variance formula: \[ \sigma_V^2 = \frac{1}{21} - \left(\frac{1}{6}\right)^2 = \frac{1}{21} - \frac{1}{36} \] Finding a common denominator (252): \[ \frac{1}{21} = \frac{12}{252}, \quad \frac{1}{36} = \frac{7}{252} \] So: \[ \sigma_V^2 = \frac{12}{252} - \frac{7}{252} = \frac{5}{252} \] --- ## **Summary of Answers** ### (a) \[ \Pr\left[\frac{1}{10} < V \leq \frac{4}{10}\right] = \boxed{0.51273} \] ### (b) - **PDF:** \[ f_V(v) = \begin{cases} 0 & v < 0 \\ 5(v-1)^4 & 0 \leq v < 1 \\ 0 & v \geq 1 \end{cases} \] - **Mean:** \[ \mu_V = \boxed{\frac{1}{6}} \] - **Variance:** \[ \sigma_V^2 = \boxed{\frac{5}{252}} \]