TMA4240 Statistikk Autumn 2013

Transcription

1 Norges teknisk-naturvitenskapelige universitet Institutt for matematiske fag Øving nummer 5, blokk I Løsningsskisse Oppgave 1 We are looking at a randomly chosen night and define the following events: A = Anne is on duty, B = Bernt is on duty, C = Cecilie is on duty, D = a case of death occurs. And assume that every death happens naturally. a Venn diagram of the four events: C B A D Since only Anne, Bernt and Cecilie have night shifts, the events A, B and C will constitute a partition of the sample space, and we must have that P (A + P (B + P (C = 1. We also see this from the venn diagram. Since Bernt and Cecilie works equally frequently, P (B = P (C. Since Anne works twice as often as each of Bernt and Cecilie, P (A = 2 P (B = 2 P (C. We express everything by P (B. P (A + P (B + P (C = 1 2 P (B + P (B + P (B = 1 P (B =.25 Therefor we have that P (A =.5 P (B =.25 P (C =.25 ov5-lsf-e. oktober 213 Side 1

2 To calculate P (D we can use the law of total probability. We know that A, B, C is a partition of the sample space. P (D = P (D A + P (D B + P (D C = P (D A P (A + P (D B P (B + P (D C P (C =.6 ( =.6 The definition of independence gives that C and D are two independent events if and only if P (D C = P (D, which means that the additional information doesn t change the situation. We see from the calculations above that P (D C = P (D =.6, and C and D is therefore independent events. Intuitively the independence of C and D follows from the assumption of naturally caused deaths. b X is a stochastic variable which describes the amount of deaths happening on Cecilie s duties, out of a total of n = 1 natural deaths. Conditions for X being binomially distributed: We look at n = 1 dødsfall. For every death we check if Cecilie was on duty or not. The probability of Cecilie being on duty given that a death has occurred is P (C D = P (C =.25, and this probability is the same for all the n deaths. The n deaths are independent since they are natural (and we assume thus that contagious diseases or epidemics are not the causes. Under these 4 conditions X= amount of natural deaths at Cecilie s duties is binomially distributed with parameters n = 1 and p =.25. Therefore the probability distribution of X is given by f(x, ( n f(x = p x (1 p n x, x =, 1,..., n x The probability that or more out of 1 deaths at night time happens on Cecilie s duties, is most easily found by using the table (page 13 in the book of formulas, P (X = = 1 P (X 6 = =.4 Let Y be a stochastic variable which represents the amount of nurses among the 3 nurses, which experiences more than deaths on their duties out of a total of 1 deaths. Y will thus be binomially distributed with n = 3 and p =.4. The probability that at least one out of the 3 nurses experiences that or more out of 1 natural deaths happen on their duties, is given by P (Y 1. ( 3 P (Y 1 = 1 P (Y = = 1.4 (1.4 3 = = 1.3 =. ov5-lsf-e. oktober 213 Side 2

3 Even if there is a small probability (only.4 that there is out of 1 deaths occurring at Cecilie s duties, there is a large probability ( percent that at least out of 1 deaths can happen on the duty of any of the nurses in norway working the same type of position as Cecilie. These observations does not strengthen the suspicion towards Cecilie. Analogy: Every week there is (usually someone who achieves right numbers in Lotto, even if this has a remarkably low probability for each Lotto-player. Oppgave 2 X and Y are independent Poisson distributed stochastic variables, X p(x;5 and Y p(y;1. From table A.2 we find P (X 5 = 5 x= p(x; 5 =.616 and P (X 3 X 5 = P (X 3 X 5 P (X 5 = P (X 3 P (X 5 =.43. Let Z = X + Y. X and Y are Poisson distributed, and therefore the variable Z will also be Poisson distributed, with parameter λ = 5+1 = 15, which means Z Poisson(15. By using table A.2 we find P (X + Y > 1 = P (Z > 1 = 1 P (Z 1 = 1 1 x= = =.882. Oppgave 3 Define X : amount of tankers (ships arriving the harbour during one day where with X poisson (λ, λ = E (X = 2. The harbor can serve a maximum of 3 tankers per day. a Because X is Poisson distributed we have that With inserted values for X we have that P (X = x = λx x! e λ = 2x x! e 2, x =, 1, 2,.... ov5-lsf-e. oktober 213 Side 3

4 x P (X = x Thus we see that, for a given day, the probability of one or two tankers is the greatest. Tankers must be directed to other harbors if more than three tankers arrive on any given day. Thus we have P (One or more tankers must be redirected = P (X > 3 = 1 P (X 3 tabell = =.1429 b Definer Y : amount of ships that is served one day. Because Y 3 we have that P (Y = y = P (X = x, y =, 1, 2 P (Y = 3 = P (X 3 = 1 P (X 2. We therefore get 3 E [Y ] = y P (Y = y y= = P (X = + 1 P (X = P (X = (1 P (X 2 = (1.66 = c Define k : capacity. We wish to find k such that P (X k.9. With inserted values for k we get Thus we have k P (X k P (X 4 =.944 >.9, and therefore we must expand the harbor to have a capacity on four ships, to be able to serve every ship arriving a given day with 9% probability. Oppgave 4 A set of lotto numbers can be regarded as an unordered selection of elements among the numbers 1 to 34, where the selection happens without replacement. ov5-lsf-e. oktober 213 Side 4

5 a There is different sets of lotto numbers. ( 34 = We are to find the probability that the set of lotto numbers contains the number 34. This can be calculated as P ( the set contains the number 34 = 1 P ( the set does NOT contain the number 34 ( 34 where the number of possible sets of lotto numbers is and the number of possible ( 33 sets of lotto numbers without the number 34 is such that ( 33 P ( the set contains the number 34 = 1 ( 34 = = 34. Define X : amount of correct numbers in the set, where X follows a Hypergeometric distribution ( ( k N k x n x h(x; N, n, k = ( N n with N = 34, n = and k =. The probability that a set of lotto numbers wil achieve correct numbers is given by ( ( 2 p = P ( correct = h(; 34,, = ( 34 = 1 ( 34 = b Every week, n = 19 sets of lotto numbers are delivered. Define X : amount out of the the n delivered sets which achieves correct numbers in this week s lotto draw ov5-lsf-e. oktober 213 Side 5

6 X is binomially distributed, but can with good approximation be regarded as Poisson distributed, because the Poisson distribution is the limiting distribution of the Binomial distribution when n and p. According to theorem 5.6 we have that n p µ when n such that the parameter in the Poisson distribution is given by µ = n p = = The probability density of X is now given by and we have f X (x = µx x! e µ P ( none of the delivered sets has correct = P (X = = 3.53 e 3.53 = e 3.53 =.29.! Observe that we get the same answer without the Poisson approximation, meaning that if X is binomially distributed with parameters n = 19 and p = given by f X (x = ( n x p x (1 p n x. We then have that P ( none of the delivered sets has correct numbers ( n = P (X = = p (1 p n = (1 p n = ( =.29. Define Y : number of Golden lotto rounds per year where we assume that there are 52 weeks per year. Y here follows a binomial distribution with n = 52 and p =.29. We therefore have that and E(Y = n p = = 1.5 P (Y = = f Y ( = ( 52 p (1 p 52 = ( =.22. ov5-lsf-e. oktober 213 Side 6

7 c Example of possible code for Poisson distribution density function: function funcval = poisson(x,mean funcval = ((meanˆx/(factorial(x*(exp( mean; TMA424 Statistikk A plot of the values from the Poisson density function for x-values from to 12 is shown in Fig 1. Here we have used µ = np = 3.53, as found in exercise 5b. We use integers, as the Poisson density function is a discrete probability distribution that is only defined for integer values of x. Figur 1: Poisson density function, f X (x = µx x! e µ for integer values of x from to 12 with µ = d Example of possible code for the binomial probability density function: function funcval = binomial(n,x,p funcval = nchoosek(n, x (pˆx ((1 pˆ(n x; A plot of the values from the binomial density function for x-values from to 12 is shown in Fig 2. Here we have used n = 19 and p = as found in exercise 5b. Figur 2: Binomial density function, f X (x = ( n x p x (1 p n x for integer values of x from to 12 with n = 19 and p = in Fig 3 we have plotted the two density functions, Poisson and binomial, against each other. We see that the density functions are approximately identical since they overlap ov5-lsf-e. oktober 213 Side

8 almost perfectly. Figur 3: Poisson density function from b plotted against the binomial density function. e We will now increase the amount of choosable numbers from 34 to m, where m > 34 such that with probability at least.1 there are no sets achieving correct numbers in a week where n = 19 sets are delivered. In this exercise we must use the equations defined earlier. We have that P ( none of the delivered sets have correct numbers = P (X = = (1 p 19.1 We have further that p is the probability that a set will achieve correct numbers, and this is given by such that p = P ( correct = h(; m,, = try with inserted values of m: 1 1 ( m ( 19 ( m k ( = m.1 m ( m 1 1 m ov5-lsf-e. oktober 213 Side 8

9 We see from the table that the inequality is satisfied for m = 36, and therefore have that m 36 if, with probability.1, there are not to exist any sets with correct numbers in a week where n = 19 sets are delivered. ov5-lsf-e. oktober 213 Side 9