Question

Find the value of \( k \).

Answer 1

Step 1: The sum of all probabilities must equal 1: \[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1.\]Step 2: Input 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: Construct the equation: \[0.1 + k(1) + k(2) + k(2) + k(1) = 1.\]Step 4: Simplify the equation: \[0.1 + 6k = 1.\]Step 5: Calculate the value of \( k \): \[k = \frac{1 - 0.1}{6} = \frac{0.9}{6} = 0.15.\]

Answer 2

Step 1: Calculate the probability of studying for at least 2 hours: \[P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4).\] Step 2: Insert values from the 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: Perform the calculation: \[P(X \geq 2) = 0.3 + 0.3 + 0.15 = 0.75.\] The probability of the student studying for at least 2 hours is 0.75.

Answer 3

Step 1: Calculate the probability of studying at most 2 hours: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2). \] Step 2: Insert given values: \[ P(X \leq 2) = 0.1 + k(1) + k(2). \] Step 3: Use \( k = 0.15 \): \[ P(X \leq 2) = 0.1 + 0.15(1) + 0.15(2). \] Step 4: Final calculation: \[ P(X \leq 2) = 0.1 + 0.15 + 0.3 = 0.55. \] The probability of a student studying at most 2 hours is 0.55.

Answer 4

Step 1: The Poisson distribution is defined by the formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]. Here, \( \lambda \) represents the average rate of occurrence, \( k \) denotes the number of occurrences, and \( e \) is the base of the natural logarithm.
Step 2: Mean expectation (\( \lambda \)): The average number of floods observed over a 10-year period is given as \( \lambda = 2 \). 
Step 3: Probability of 3 or fewer overflows (\( P(X \leq 3) \)): This is calculated by summing the probabilities of 0, 1, 2, and 3 overflows: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3). \] Applying the Poisson formula for each case: - 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: Summing the individual probabilities yields the total probability: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3). \] Substituting the calculated 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: \( P(X \leq 3) = 0.85715 \), which is approximately \( 85.72\% \).