Question

Find the value of \( k \).

Answer 1

Step 1: The total probability must sum to 1: \[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1. \] Step 2: Substitute the given probability values: \[ P(X = 0) = 0.1, \quad P(X = 1) = k(1), \quad P(X = 2) = k(2), \] \[ P(X = 3) = k(5 - 3), \quad P(X = 4) = k(5 - 4). \] Step 3: Form the equation: \[ 0.1 + k(1) + k(2) + k(2) + k(1) = 1. \] Step 4: Simplify the equation: \[ 0.1 + 6k = 1. \] Step 5: Solve for \( k \): \[ k = \frac{1 - 0.1}{6} = \frac{0.9}{6} = 0.15. \]

Answer 2

Step 1: The probability of the student studying for at least 2 hours is given by: \[ P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4). \] Step 2: Substitute the values from the given probability distribution: \[ P(X \geq 2) = k(2) + k(5 - 3) + k(5 - 4). \] Step 3: Substitute \( k = 0.15 \): \[ P(X \geq 2) = 0.15(2) + 0.15(2) + 0.15(1). \] Step 4: Compute the value: \[ P(X \geq 2) = 0.3 + 0.3 + 0.15 = 0.75. \] Thus, the probability that the student studied for at least 2 hours is **0.75**.

Answer 3

Step 1: The probability that the student studied for at most 2 hours is given by: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2). \] Step 2: Substitute the given values: \[ P(X \leq 2) = 0.1 + k(1) + k(2). \] Step 3: Substitute \( k = 0.15 \): \[ P(X \leq 2) = 0.1 + 0.15(1) + 0.15(2). \] Step 4: Compute the value: \[ P(X \leq 2) = 0.1 + 0.15 + 0.3 = 0.55. \] Thus, the probability that the student studied for at most 2 hours is 0.55.

Answer 4

Step 1: The Poisson distribution formula is: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \] where \( \lambda \) is the average rate of occurrence, \( k \) is the number of occurrences, and \( e \) is the base of the natural logarithm. 
Step 2: Mean expectation (\( \lambda \)): The average number of floods in 10 years is \( \lambda = 2 \). 
Step 3: Probability of 3 or fewer overflows (\( P(X \leq 3) \)): \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3). \] Using the Poisson formula: - For \( P(X = 0) \): \[ P(X = 0) = \frac{2^0 e^{-2}}{0!} = \frac{1 \cdot 0.13534}{1} = 0.13534. \] - For \( P(X = 1) \): \[ P(X = 1) = \frac{2^1 e^{-2}}{1!} = \frac{2 \cdot 0.13534}{1} = 0.27068. \] - For \( P(X = 2) \): \[ P(X = 2) = \frac{2^2 e^{-2}}{2!} = \frac{4 \cdot 0.13534}{2} = 0.27068. \] - For \( P(X = 3) \): \[ P(X = 3) = \frac{2^3 e^{-2}}{3!} = \frac{8 \cdot 0.13534}{6} = 0.18045. \] 
Step 4: Add the probabilities: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3). \] Substitute the values: \[ P(X \leq 3) = 0.13534 + 0.27068 + 0.27068 + 0.18045 = 0.85715. \] 
Final Answers: - Mean expectation: \( \lambda = 2 \). - Probability of 3 or fewer overflows: \( P(X \leq 3) = 0.85715 \) or approximately \( 85.72\% \).