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İstanbul Kültür University Faculty of Engineering MCB17 Introduction to Probability Statistics Second Midterm Fall 21-21 Number: Name: Department: Section: Directions You have 9 minutes to complete

1 İstanbul Kültür University Faculty of Engineering MCB17 Introduction to Probability Statistics Second Midterm Fall Number: Name: Department: Section: Directions You have 9 minutes to complete the exam. Please do not leave the examination room in the first minutes of the exam. There are six questions, of varying credit (1 points total). Indicate clearly your final answer to each question. You are allowed to use a calculator. During the exam, please turn off your cell phone(s). You cannot use the book or your notes. You have one page for cheat-sheet notes. The answer key to this exam will be posted on Department of Mathematics Computer Science board after the exam. Good luck! Emel Yavuz Duman, PhD. M. Fatih Uçar, PhD. Question 1. Question. Question 2. Question 5. Question. Question 6. TOTAL

2 Queflion points Suppose that a bag has boxes, one of them containing 2 bullets two of them containing 1 bullet. A person is asked to romly choose a box than shoot a target until the last bullet is fired. Suppose 9% for this person, the bullet will hit the target. Let X be the number of bullets hitting the target Y be the number of bullets in the selected box that the person has. (a) Find the joint probability distribution of X Y. Answer. For the values of X =, 1, 2 Y =1, 2 we have the following probabilities: f(, 1) = 2 ( ) = 1 15 = 2, f(, 2) = 1 ( ) = 1 f(1, 1) = 2 ( ) = 1 5 = 18, f(1, 2) = 1 ( ) = 1 5 = 18 f(2, 1) =, f(2, 2) = 1 ( ) = = 81 Therefore, the joint probability distribution of X Y is x y 1 2 h(y) 1 2/ 18/ 2/ 2 1/ 18/ 81/ 1/ g(x) 21/ 198/ 81/ 1 (b) Find the conditional variance of X given Y =1. Answer. Conditional distribution of X given Y =1is f(x, 1) f(x 1) = = f(, 1) = 2 = 1 for x = f(x, 1) = h(1) 2 2 f(1, 1) = 18 = 9 for x =1 2 1 f(2, 1) = = for x =2 2 2 On the other h, since E[X 1] = 2 xf(x 1) = f( 1) + 1f(1 1) + 2f(2 1) = 9 1 x= E[X 2 1] = 2 x= Thus the conditional variance of X given Y =1is x 2 f(x 1) = 2 f( 1) f(1 1) f(2 1) = 9 1 σ 2 X 1 = E[(X μ X 1) 2 1] = E[X 2 1] (E[X 1]) 2 = 9 1 ( 9 1 ) 2 = 9 1. MCB17 - Int. to Prob. Statistics 2 Second Midterm

3 Queflion points (a) The moment-generating function of the rom variable X is M X (t). Ifwelet R X (t) =lnm X (t), show that d R dt X(t) = μ d2 R dt 2 X (t) = σ 2. t= t= ( ) d Answer. Hint: ln u(x) = u (x) dx u(x) It is easy to see that d dt R X(t) = d t= dt ln M X(t) = M X (t) = M X () t= M X (t) t= M X () = μ 1 1 = μ, d 2 dt R X(t) = d 2 t= dt [ ] M X (t) = M X (t)m X(t) (M X (t))2 t= M X (t) t= (M X (t)) 2 = M X ()M X() (M X ())2 = μ 2 μ 2 (M X ()) 2 1 = σ 2. (b) Use the results given in (a) to find the mean the variance of X having the momentgenerating function M X (t) =e (et 1). Answer. Since M X (t) =e (et 1) R X (t) = log M X (t) =(e t 1) then, we see that μ = d dt ((et 1)) =e t t= =, t= σ 2 = d2 dt 2 ((et 1)) = e t t= =. MCB17 - Int. to Prob. Statistics Second Midterm

4 Queflion. A rom variable X has the density function given by { 6x(1 x) for <x<1 f(x) = elsewhere. 8+7points (a) Find P ( X μ.7). Answer. If we use the definition of a mean, we obtain that μ = E[X] = ( x x On the other h, since ) 1 xf(x)dx = ( 1 1 6x 2 (1 x)dx ) = 1 2 =.5. (x 2 x )dx P ( X μ.7) = P ( X.5.7) = 1 P ( X.5 <.7), X.5 <.7.7 <X.5 < <X< <X<1.2, thus P ( X.5.7) = 1 P ( X.5 <.7) = f(x)dx =1 f(x)dx =1 1 =. (b) Find an upper bound for P ( X.5.7) compare it with the result in (a). Answer. We need to use the Chebyshev s inequality given by P ( X μ ε) σ 2 /ε 2 where mean μ = E(X) =.5 ε =.7. VarianceofX can be calculated by using the formula σ 2 = E(X 2 ) [E(X)] 2.Since E[X 2 ]= x 2 f(x)dx = ( ) x 1 x5 6x (1 x)dx ( ) = 1 =. then σ 2 = E(X 2 ) [E(X)] 2 =..5 2 =.5. So, an upper bound for the given probability is (x x )dx P ( X.5.7).5.7 = Therefore, we can say that an upper bound for the given probability is.12 which is quite far from the real value which is. MCB17 - Int. to Prob. Statistics Second Midterm

5 Queflion points A doctor wishes to perform an experiment on 5 patients. Suppose that the probability that a romly selected patient agrees to participate to this experiment is.2. (a) What is the probability that 15 patients must be asked before 5 are found who agree to participate? Answer. X, the number of couples who agree to participate, is a negative-binomial rom variable. So, ( ) 1 b (15; 5,.2) = (.2) 5 (.8) 1 =.9 (b) What is the probability that at least patients must be asked before the first patient is found who agree to participate? Answer. X, the number of couples who agree to participate, is a geometric rom variable. So, P (X ) = 1 P (X <) = 1 [g(1;.2) + g(2;.2) + g(;.2)] =1 [ (.2)(.8) +(.2)(.8) 1 +(.2)(.8) 2] =1.88 =.512 Queflion points Suppose you roll a pair of non-fair, six sided dice 1 times, which are loaded in such a way that each odd number is three times as likely to occur as each even number. Let the rom variable X denote the sum of the points on both dice which can come up in one roll. Use a Poisson approximation to determine the probability of observing that X exactly four times in this 1 rolls. Answer. Since S = {1, 2,,, 5, 6}, each odd number is three times as likely to occuraseachevennumber,wehave P (X =2)=P (X =)=P (X )=p P (X =1)=P (X =)=P (X =5)=p. So, we get P (X =1)+P (X =2)+P (X =)+P (X =)+P (X =5)+P (X )=12p =1 p = Let we define A = {sum of the points on both dice which can come up is 6}. Thus,we have A = {(, 2), (2, ), (5, 1), (1, 5), (, )}. Therefore, we get P (A) = = 29 1 = θ. Hence, using Poisson approximation we obtain that the probability of observing that X exactly four times in the 1 rolls with the parameter λ = nθ = 1 29 =29is 1 p (; 29) = 29 e 29! = MCB17 - Int. to Prob. Statistics 5 Second Midterm

6 Queflion points Suppose that there are two types of airplanes, one of them has two engines other one has four engines. It is given that an airplane engine will fail, during the flight, with probability 1 θ (where θ 1), independently from engine to engine; suppose that the airplane will make a successful flight if at least 5 percent of its engines remain operative. For what values of θ is a two-engine plane preferable to a four-engine plane? Answer. As each engine is assumed to fail or function independently of what happens with the other engines, it follows that X, the number of engines remaining operative, is a binomial rom variable. Hence, the probability that a four-engine plane makes a successful flight is P (X 2) = 1 P (X <2) = 1 (P (X =)+P (X =1)) (( ) ( ) =1 θ (1 θ) + )θ 1 (1 θ) 1 1 =1 (1 θ) θ(1 θ) = whereas the corresponding probability for a two-engine plane is ( ) 2 P (X 1) = 1 P (X <1) = 1 P (X =)=1 θ (1 θ) 2 =1 (1 θ) 2 Hence, the two-engine plane is safer if 1 (1 θ) 2 1 (1 θ) θ(1 θ) (1 θ) 2 (1 θ) +θ(1 θ) 1 (1 θ) 2 +θ(1 θ) =1 2θ + θ 2 +θ θ 2 2θ θ 2 2θ θ 2 = θ(2 θ) 2 θ θ 2 MCB17 - Int. to Prob. Statistics 6 Second Midterm

MCB1007 Introduction to Probability and Statistics. First Midterm. Fall Solutions

MCB1007 Introduction to Probability and Statistics. First Midterm. Fall Solutions İstanbul Kültür University MCB7 Introduction to Probability and Statistics First Midterm Fall 4-5 Solutions Directions You have 9 minutes to complete the eam Please do not leave the eamination room in