A polling station has a mean arrival rate of 5 voters per 20 minutes. Using the Poisson distribution, calculate the probability of (a) no voters arriving between 3:10 pm and 3:30 pm, and (b) exactly 5

Welcome back — missed you this week. Here are the solutions to the problem: Part (i): Poisson Distribution The number of voters arriving at a polling station follows a Poisson distribution with a rate of = 5 voters per 20 minutes. The probability mass function for a Poisson distribution is P(X=k) = e^- ^kk!. a) No voter will arrive between 3:10 and 3:30 pm Step 1: Determine the time interval. The interval is 3:30 - 3:10 = 20 minutes. Step 2: Identify the rate parameter for this interval. For a 20-minute interval, = 5. Step 3: Calculate the probability for k=0. P(X=0) = e^-5 5^00! = e^-5 · 11 = e^-5 P(X=0) ≈ 0.0067379 The probability that no voter will arrive is 0.0067. b) There will be exactly 5 voters between 9:20-10:20 Step 1: Determine the time interval. The interval is 10:20 - 9:20 = 60 minutes. Step 2: Calculate the rate parameter ' for this new interval. Since the rate is 5 voters per 20 minutes, for 60 minutes (which is 3 × 20 minutes), the rate is: ' = 5 × (60)/(20) = 5 × 3 = 15 Step 3: Calculate the probability for k=5 with ' = 15. 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 of exactly 5 voters is 0.0019. Part (ii): Geometric Distribution The probability of getting a wife (success) on any given church service is p = (1)/(5). The number of times he attends church service until he gets a wife follows a Geometric distribution, P(X=k) = (1-p)^k-1p. c) The mean and variance of the number of times he attends church service Step 1: State the probability of success. p = (1)/(5) = 0.2. Step 2: Calculate the mean E(X). E(X) = (1)/(p) = (1)/(0.2) = 5 Step 3: Calculate the variance Var(X). Var(X) = (1-p)/(p^2) = (1-0.2)/((0.2)^2) = (0.8)/(0.04) = 20 The mean is 5 and the variance is 20. d) The probability that he finds a wife on the 5th church service Step 1: Identify the number of trials k. We need the probability that X=5. Step 2: Apply the Geometric probability formula. P(X=5) = (1-p)^5-1p = (1-(1)/(5))^4 × (1)/(5) P(X=5) = ((4)/(5))^4 × (1)/(5) = (256)/(625) × (1)/(5) = (256)/(3125) P(X=5) = 0.08192 The probability is 0.0819. 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\% = 0.01. So, (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 natural logarithm of both sides: n (0.8) < (0.01) Since (0.8) is negative, divide by (0.8) and reverse the inequality sign: n > ((0.01))/((0.8)) n > (-4.60517)/(-0.22314) n > 20.638 Step 3: Determine the least integer n. The least integer n that satisfies n > 20.638 is 21. The least integer n is 21. Send me the next one 📸