a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf

a) mean = 1.4*3 Thus distribution is Poisson(4.2) P(X = 4) = 4.2^4 * e^{-4.2} / 4! = 0.1944 b) average number of ghosts in 90min (1.5hrs) interval = 1.4*1.5 = 2.1 c) Probability that 4th ghost will appear before 10 am = Probability that number of ghosts appearing from 8 to 10 am >= 4 As it is a 2 hr period, the distribution of number of ghosts is Poisson(1.4*2) i.e Poisson(2.8) P(X >= 4) = 1 - P(X < 4) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 - [(2.8^0 * e^{-2.8} / 0!) + (2.8^1 * e^{-2.8} / 1!) + (2.8^2 * e^{-2.8} / 2!) + (2.8^3 * e^{- 2.8} / 3!)] = 1 - [0.0608 + 0.1703 + 0.2384 + 0.2225] = 0.692 d) When arrival distribution is Poisson(lambda), inter-arrival time distribution is Exponential(lambda) Thus, interarrival time distribution is Exponential(1.4) Expected value of time between 5th and 6th ghost = 1/1.4 = 0.7143hr e) On average time taken for 10th ghost to arrive = 10*(1/1.4) = 7.143hr Thus on an average, 10th ghost will arrive at 3:09pm f) Expected value of time between ghost appearances = 1/1.4 = 0.7143hr (see part d) g) Probability that 3rd ghost appears after 9:45 am = Probablity that number of ghosts appearing till 9:45 am < 3 Duration from 8am to 9:45 am = 1.75hr Mean number of ghosts = 1.4*1.75 = 2.45 Thus X is Poisson(2.45) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = (2.45^0 * e^{-2.45} / 0!) + (2.45^1 * e^{-2.45} / 1!) + (2.45^2 * e^{-2.45} / 2!) = 0.0863 + 0.2114 + 0.2590 = 0.5567 h) As inter-arrival time is Exponential(1.4), inter-arrival time is memoryless Thus, time for 7th ghost will be calculated from 1:00 pm and not from 12:35pm due to memorylessness property. Time duration from 1:00pm to 1:15pm = 0.25hrs Let Y be random variable denoting inter-arrival time probability that the 7th ghost will appear before 1:15 p.m. = P(Y < 0.25) = 1 - e^{-1.4 * 0.25) = 0.2953 i) Expected value of time for 7th ghost to appear after 1:00pm = 1/1.4 = 0.7143hr = 42.86 min Time from 12:35pm to 1:00pm = 25min Thus expected interarrival time = 25 + 42.86 = 67.86 min = 1hr 7.86min j) Expected duration after 1:00pm for 9th ghost to appear = 3 * (1/1.4) = 2.143hr = 2hr 8.58min Thus expected time at which 9th ghost appears = 3:09pm k) As the inter-arrival time follows memorylessness property, the future probabilities do not change with the observed value at past. Thus, 4 ghosts appearing between 7:00pm to 10:00pm does not affect ghost appearance from 11:00pm to 11:30pm Duration from 11:00pm to 11:30pm = 0.5hrs mean number of ghosts = 1.4*0.5 = 0.7 Thus it follows Poisson(0.7) distribution P(X > 2) = 1 - P(X <= 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] = 1 - [(0.7^0 * e^{-0.7} / 0!) + (0.7^1 * e^{-0.7} / 1!) + (0.7^2 * e^{-0.7} / 2!)] = 1 - [0.4966 + 0.3476 + 0.1217] = 0.0341 l) Probability that 8 ghosts appear between 6pm and midnight = Probability that number of ghosts appear from 6pm to 7pm and 10pm to midnight = 4 as it is known that 4 ghosts appeared from 7pm to 10pm Let A denote number of ghost from 6pm to 7pm B denote number of ghosts .Read less