This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Hey, good to see you again. Part (i): Poisson Distribution The number of voters arriving at a polling station follows a Poisson distribution. The average rate is given as _0 = 5 voters in 20 minutes. The probability mass function for a Poisson distribution is P(X=k) = e^- ^kk!. a) Calculate the probability that no voter will arrive between 3:10 and 3:30 pm. Step 1: Determine the time interval and corresponding . The time interval is from 3:10 pm to 3:30 pm, which is 3:30 - 3:10 = 20 minutes. For a 20-minute interval, the average rate = 5. Step 2: Calculate the probability for X=0. We want to find P(X=0). P(X=0) = e^-5 5^00! = e^-5 · 11 = e^-5 P(X=0) ≈ 0.006738 The probability that no voter will arrive is 0.006738. b) Calculate the probability that there will be exactly 5 voters between 9:20-10:20. Step 1: Determine the time interval and corresponding . The time interval is from 9:20 to 10:20, which is 10:20 - 9:20 = 60 minutes. Since the rate is 5 voters per 20 minutes, for a 60-minute interval, the new average rate ' is: ' = 5 voters20 minutes × 60 minutes = 5 × 3 = 15 Step 2: Calculate the probability for X=5 with ' = 15. We want to find P(X=5). P(X=5) = e^-15 15^55! P(X=5) = e^-15 × 759375120 P(X=5) ≈ 3.05902 × 10^-7 × 759375120 P(X=5) ≈ (0.23226)/(120) ≈ 0.0019355 The probability that there will be exactly 5 voters is 0.001936. Part (ii): Geometric Distribution Let X be the number of church services attended until the first success (finding a wife). The probability of getting a wife (success) on any given service is p = (1)/(5). The probability mass function for a Geometric distribution (number of trials until first success) is P(X=k) = (1-p)^k-1p. c) Calculate the mean and variance of the number of times he attends church service. Step 1: Calculate the mean. For a Geometric distribution, the mean E(X) is given by: E(X) = (1)/(p) E(X) = (1)/(1/5) = 5 Step 2: Calculate the variance. For a Geometric distribution, the variance Var(X) is given by: Var(X) = (1-p)/(p^2) Var(X) = (1 - 1/5)/((1/5)^2) = (4/5)/(1/25) = (4)/(5) × 25 = 20 The mean is 5 and the variance is 20. d) Calculate the probability that he finds a wife on the 5th church service. Step 1: Use the Geometric probability formula for X=5. We want to find P(X=5). P(X=5) = (1-p)^5-1p P(X=5) = (1 - (1)/(5))^4 × (1)/(5) P(X=5) = ((4)/(5))^4 × (1)/(5) P(X=5) = (256)/(625) × (1)/(5) = (256)/(3125) P(X=5) = 0.08192 The probability that he finds a wife on the 5th church service is 0.08192. e) Find the least integer n, the number of times he attends church service such that P(X > n) < 1\%. Step 1: Set up the inequality for P(X > n). For a Geometric distribution, P(X > n) = (1-p)^n. We are given P(X > n) < 1\%, which is P(X > n) < 0.01. So, we need to solve: (1 - (1)/(5))^n < 0.01 ((4)/(5))^n < 0.01 0.8^n < 0.01 Step 2: Solve for n using logarithms. Take the logarithm (base 10 or natural log) of both sides. Using _10: n _10(0.8) < _10(0.01) Calculate the logarithm values: _10(0.8) ≈ -0.09691 _10(0.01) = -2 Substitute these values into the inequality: n (-0.09691) < -2 Divide by -0.09691. Remember to reverse the inequality sign when dividing by a negative number: n > (-2)/(-0.09691) n > 20.637 Step 3: Determine the least integer n. Since n must be an integer and n > 20.637, the least integer value for n is 21. The least integer n is 21. Send me the next one 📸