Given the below cumulative distribution function, find the following probabilities. CDF: \( F(x)=\left\{\begin{array}{ll}0 & \text { if } x<3 \\ 0.15 & \text { if } 3 \leq x<5 \\ 0.3 & \text { if } 5 \leq x<10 \\ 0.35 & \text { if } 10 \leq x<11 \\ 1 & \text { if } 11 \leq x\end{array}\right. \) \( P(X=0)= \) \( P(X \leq 5)= \) \( P(X<10)=\square \) \( P(10 \leq X \leq 11)=\square \) \( P(310)= \)

Let's analyze the given cumulative distribution function (CDF) to determine the required probabilities. **Given CDF:** \[ F(x) = \begin{cases} 0 & \text{if } x < 3 \\ 0.15 & \text{if } 3 \leq x < 5 \\ 0.3 & \text{if } 5 \leq x < 10 \\ 0.35 & \text{if } 10 \leq x < 11 \\ 1 & \text{if } 11 \leq x \end{cases} \] From the CDF, we can deduce the probability mass function (PMF) for the discrete random variable \( X \): \[ \begin{align*} P(X = 3) &= F(3) - F(3^-) = 0.15 - 0 = 0.15 \\ P(X = 5) &= F(5) - F(5^-) = 0.3 - 0.15 = 0.15 \\ P(X = 10) &= F(10) - F(10^-) = 0.35 - 0.3 = 0.05 \\ P(X = 11) &= F(11) - F(11^-) = 1 - 0.35 = 0.65 \\ \end{align*} \] Assuming \( X \) only takes the values 3, 5, 10, and 11, let's compute the requested probabilities: 1. **\( P(X = 0) \)** Since \( X \) takes values starting from 3: \[ P(X = 0) = 0 \] 2. **\( P(X \leq 5) \)** \[ P(X \leq 5) = P(X = 3) + P(X = 5) = 0.15 + 0.15 = 0.3 \] 3. **\( P(X < 10) \)** \[ P(X < 10) = P(X = 3) + P(X = 5) = 0.15 + 0.15 = 0.3 \] 4. **\( P(10 \leq X \leq 11) \)** \[ P(10 \leq X \leq 11) = P(X = 10) + P(X = 11) = 0.05 + 0.65 = 0.7 \] 5. **\( P(3 < X \leq 10) \)** \[ P(3 < X \leq 10) = P(X = 5) + P(X = 10) = 0.15 + 0.05 = 0.2 \] 6. **\( P(3 < X < 11) \)** \[ P(3 < X < 11) = P(X = 5) + P(X = 10) = 0.15 + 0.05 = 0.2 \] 7. **\( P(X \geq 5) \)** \[ P(X \geq 5) = P(X = 5) + P(X = 10) + P(X = 11) = 0.15 + 0.05 + 0.65 = 0.85 \] 8. **\( P(X > 10) \)** \[ P(X > 10) = P(X = 11) = 0.65 \] **Summary of Probabilities:** \[ \begin{align*} P(X = 0) &= 0 \\ P(X \leq 5) &= 0.3 \\ P(X < 10) &= 0.3 \\ P(10 \leq X \leq 11) &= 0.7 \\ P(3 < X \leq 10) &= 0.2 \\ P(3 < X < 11) &= 0.2 \\ P(X \geq 5) &= 0.85 \\ P(X > 10) &= 0.65 \\ \end{align*} \]