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HW506Sol3 Washington STAT 506
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  • Title: HW506Sol3
  • Type: Notes
  • School: Washington
  • Course: STAT 506
  • Term: Spring

Coursehero >> Washington >> Washington >> STAT 506
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506, Stat Homework set #3 Due Monday April 21, 2008 From Casella and Berger. 3.3; 3.7; 3.9; 3.20 and 3.23 Solution to Casella and Berger 3.3 Let Xi be the indicator function of the event a car is passing during the i-th second , where we start counting time when the pedestrian arrives. The assumptions given imply that X1 , X2 , . . . is a sequence of Bernoulli trials with parameter p. We are also told that the pedestrian will take 3 seconds to cross. We are asked for the probability that the pedestrian has to wait 4 seconds before starting to cross. There is information missing in the statement of this problem. Do we want the probability that the pedestrian makes a safe crossing? Can the pedestrian see whether a car is coming and so he will know that a car will be crossing a few seconds from now? If the answer to this last question is yes, how far can the pedestrian see? If the pedestrian can t see what is coming or we don t care whether the pedestrian makes a safe crossing then the probability is p4 . On the other hand if the pedestrian can see far enough for incoming cars, or if we care for a safe crossing, then the probability is p4 (1 p)3 . Solution to Casella and Berger 3.7 X, which denotes the number of chips, is a Poisson random variable with parameter . Remember that for a Poisson random variable E(X) = . We want the smallest such that P (X 2) .99, or equivalently the smallest such that P (X < 2) .01. Since P (X < 2) = P (X = 0) + P (X = 1) = (1 + )e , then we want the smallest such that (1 + )e .01. The function (1 + )e is a decreasing function of and so the solution we are looking for is the root of the equation (1+ )e = .01. Using some numerical method, e.g., bisection, secant or Newton-Raphson, we get that 6.64. Solution to Casella and Berger 3.9 (a) In the rst paragraph of this problem it states ... reported that its incoming kindergarten class contained ve sets of twins. The model suggested in part (a) for X, the number of twins in a group of 60 students, is that of a binomial random variable with parameters n = 60 1 and p = 1/90. This model is not consistent with the above quote. The binomial model assumes independence of all 60 students. On the other hand the quote talks about sets of twins, not about individual children; don we t have independence. Nevertheless let us solve this problem accepting the binomial model. Then 60 P (X 5) = i=5 4 60 i 60 i p (1 p)60 i = 1 p (1 p)60 i 0.000556628 i i i=0 One then would say that this is a rare event. (b) Having accepted the above calculation let Y denote the number of schools in New York state that have at least 5 twins. Now we may ask the question, what is the chance that at least one school will have at least 5 twins? The presence of twins may be considered independent from school to school, so it is reasonable to model Y as binomial with parameters n = 310 and p = P (X 5) 0.000556628: P (Y 1) = 1 (1 p)310 0.1585281 i.e., almost a 16% chance. (c) For this last case we increase n to about 15000 and so P (Y 1) 0.999764, i.e., an event almost sure to occur. What is the di erence between the meaning of the probabilities that we obtained in (a), (b) and (c)? The rst one would apply if for a given xed school we asked what is the probability that in THAT school we nd 5 children that are twins? . For the last two cases we are asking what is the probability that in some xed district school, and for some de ned period of time, there is at least one school which at some time had 5 twins? The probability computed in part (c) shows that is very likely to occur; its occurrence should not surprise anybody. Solution to Casella and Berger 3.20 (a) See solution to Problem 2.11 part (b). (b) The gamma density is cx 1 e x where c is a constant. Then we need 2 a transformation that gets rid of the square in the part e x of the density f (x). We can try Y = X 2 /2 (we could use any multiple), which give a Gamma distribution with parameters = 1 and = 1/2. Solution to Casella and Berger 3.23 2 Let f (x) = x ( +1) for x > (where , > 0). (a) First f (x) 0 for all x since an exponential function is always positive. f (x) dx = x ( +1) dx = x =1 (b) and (c) We can consider all moments: E(X n ) = x ( +1 n) dx This integral converges only if n < and for those cases we get E(X n ) = n /( n). In particular, if > 2, then E(X) = /( 1), and E(X 2 ) = 2 /( 2). Then use Var(X) = E(X 2 ) E 2 (X). 3

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