1) Problem: Given the piecewise function \( f(x) = \begin{cases} cx^2, & 0 < x < 3 \\ 0, & \text{elsewhere} \end{cases} \), find the constant \( c \) such that \( f(x) \) is a density function, and compute probabilities \( P(1 < x < 2) \), \( P(x=0) \), and \( P(x > 2) \). 2) Problem: A random variable \( X \) has a Poisson distribution with mean 3. Find the probability that \( X \) is bounded by 1 and 3. 3) Problem: Determine the cumulative density function (CDF) of a random variable with an exponential distribution having variance \( 2s \). 4) Problem: Given the pmf \( f(x) = \begin{cases} \frac{x+1}{15}, & x=0,1,2,3,4 \\ 0, & \text{elsewhere} \end{cases} \), find the pmf of \( Y = (X-2)^2 \). --- ### Step 1: Find \( c \) for density function 1. The total probability must be 1, so \( \int_{-\infty}^{\infty} f(x) dx = 1 \). 2. Since \( f(x) = cx^2 \) for \( 0 < x < 3 \) and 0 elsewhere, $$ \int_0^3 cx^2 dx = 1 $$ 3. Compute the integral: $$ c \int_0^3 x^2 dx = c \left[ \frac{x^3}{3} \right]_0^3 = c \cdot \frac{27}{3} = 9c $$ 4. Set equal to 1: $$ 9c = 1 \implies c = \frac{1}{9} $$ --- ### Step 2: Compute probabilities - \( P(1 < x < 2) = \int_1^2 f(x) dx = \int_1^2 \frac{1}{9} x^2 dx = \frac{1}{9} \left[ \frac{x^3}{3} \right]_1^2 = \frac{1}{9} \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{7}{27} \). - \( P(x=0) = 0 \) since continuous distributions have zero probability at a point. - \( P(x > 2) = \int_2^3 \frac{1}{9} x^2 dx = \frac{1}{9} \left[ \frac{x^3}{3} \right]_2^3 = \frac{1}{9} \left( 9 - \frac{8}{3} \right) = \frac{19}{27} \). --- ### Step 3: Poisson distribution probability 1. \( X \sim \text{Poisson}(\lambda=3) \). 2. Probability \( P(1 \leq X \leq 3) = P(X=1) + P(X=2) + P(X=3) \). 3. Poisson pmf: $$ P(X=k) = \frac{e^{-3} 3^k}{k!} $$ 4. Calculate each term: $$ P(X=1) = e^{-3} \frac{3^1}{1!} = 3e^{-3} $$ $$ P(X=2) = e^{-3} \frac{3^2}{2!} = \frac{9}{2} e^{-3} $$ $$ P(X=3) = e^{-3} \frac{3^3}{3!} = \frac{27}{6} e^{-3} = \frac{9}{2} e^{-3} $$ 5. Sum: $$ P(1 \leq X \leq 3) = e^{-3} \left( 3 + \frac{9}{2} + \frac{9}{2} \right) = e^{-3} (3 + 9) = 12 e^{-3} $$ --- ### Step 4: CDF of exponential distribution with variance \( 2s \) 1. Variance of exponential distribution is \( \sigma^2 = \frac{1}{\lambda^2} \). 2. Given \( \sigma^2 = 2s \), so: $$ \frac{1}{\lambda^2} = 2s \implies \lambda = \frac{1}{\sqrt{2s}} $$ 3. The CDF of exponential distribution is: $$ F(x) = 1 - e^{-\lambda x}, \quad x \geq 0 $$ 4. Substitute \( \lambda \): $$ F(x) = 1 - e^{-\frac{x}{\sqrt{2s}}} $$ --- ### Step 5: PMF of \( Y = (X-2)^2 \) 1. Possible values of \( X \) are 0,1,2,3,4. 2. Compute \( Y \) values: - \( X=0 \Rightarrow Y = (0-2)^2 = 4 \) - \( X=1 \Rightarrow Y = (1-2)^2 = 1 \) - \( X=2 \Rightarrow Y = (2-2)^2 = 0 \) - \( X=3 \Rightarrow Y = (3-2)^2 = 1 \) - \( X=4 \Rightarrow Y = (4-2)^2 = 4 \) 3. Group probabilities by \( Y \) values: - \( Y=0 \): from \( X=2 \), \( P(Y=0) = P(X=2) = \frac{2+1}{15} = \frac{3}{15} = \frac{1}{5} \) - \( Y=1 \): from \( X=1,3 \), sum probabilities: $$ P(Y=1) = P(X=1) + P(X=3) = \frac{1+1}{15} + \frac{3+1}{15} = \frac{2}{15} + \frac{4}{15} = \frac{6}{15} = \frac{2}{5} $$ - \( Y=4 \): from \( X=0,4 \), sum probabilities: $$ P(Y=4) = P(X=0) + P(X=4) = \frac{0+1}{15} + \frac{4+1}{15} = \frac{1}{15} + \frac{5}{15} = \frac{6}{15} = \frac{2}{5} $$ 4. Final pmf of \( Y \): $$ f_Y(y) = \begin{cases} \frac{1}{5}, & y=0 \\ \frac{2}{5}, & y=1 \\ \frac{2}{5}, & y=4 \\ 0, & \text{elsewhere} \end{cases} $$