BLAUGRANA LIVE now Notification 17. Let Y have a Poisson distribution with mean A. FindE[Y(Y – 1)] and then use this to show thatVar(Y) = 18. A salesperson has found that the probability of a sale on a single contact is approximately .03. If the salesperson contacts 100 prospects, what is the approximate probability of making at least one sale? 19. Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. During a given hour, what are the probabilities that (a) no more than three customers arrive? (b) at least two customers arrive? (c) exactly four customers arrive? 20. (a) The random variable Y has a Poisson distribution and is such that P(Y = 0) = P(Y = 1). What is P(Y2 = 1)? (b) Cars arrive at a toll both according to a Poisson process with mean 80 cars per hour. If the attendant makes a one-minute phone call, what is the probability that at least 1 car arrives during the call? 21. The mean of nobiles entering a mountain tunnel per two-minute period is one. An excessive number of cars entering the tunnel during a brief period of time produces a hazardous situation. Find the probability that the number of autos entering the tunnel during a two-minute period exceeds three. Does the Poisson model seem reasonable for this problem? 22. The number of claims per month paid by an insurance company is modelled by a random variable N with p.m.f satisfying the relation p(n + 1) = n = 0,1, 2, ... where p(n) is the probability that n claims are filed during a given month (a) Find p(0). (b) Calculate the probability of at least one claim during a particular month given that there have been at most four claims during the month. 23. Let Y denote a geometric random variable with probability of success p. (a) Show that for a positive integer a, P(Y > a) = (1 – p)ª (b) Show that for a positive integer a and b, P(Y > a + b|Y > a) = (1 – p)ª = P(Y > b). (c) Why do you think the property in (b) is called the memoryless property of the geometric distribution? (d) In the development of the distribution of the geometric random variable, we assumed that the experiment consisted of conducting identical and independent trials until the first success was observed. In light of these assumptions, why is the result in part (b) "obvious"? (e) Show that P(Y = an odd integer) = 1- q? 24. Given that the random variable W is binomial distribution with n trials and success probability p in each trial and P(W = w) = h(w), show that STAT 221/Exx1 Page 3 of 4 IS/AL/EA 2021 h(w) (a) h(w – 1) p(n – w + 1) (1 – p)w , w>0 (b) E[W(W – 1)]= n(n – 1)p?. E W 1- (1– p)"+1 (n + 1)p +1
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
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