English

not defined

no text concepts found

İstanbul Kültür University Faculty of Engineering MCB1007 Introduction to Probability and Statistics Second Midterm Fall 2014-2015 Number: Name: Department: – You have 90 minutes to complete the exam. Please do not leave the examination room in the ﬁrst 30 minutes of the exam. There are six questions, of varying credit (100 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 at the end of the exam papers. Good luck! Question 1. Question 2. Question 3. Emel Yavuz Duman, PhD. Question 4. Question 5. Question 6. TOTAL 17 points A statistics professor classiﬁes his students according to their grade point average (GPA) and their gender. The accompanying table gives the proportion of students falling into the various categories. PP PP GPA P Gender PPPP Under 2.0 2.0 − 3.0 Over 3.0 0.05 0.10 0.25 0.30 0.10 0.20 Male Female One student is selected at random. Let we deﬁne two random variables X and Y as ⎧ ⎪ ⎨0, student has GPA under 2.0 0, student is a male X= , Y = 1, student has GPA between 2.0 and 3.0 ⎪ 1, student is a female ⎩ 2, student has GPA over 3.0 Find the conditional variance of Y given that X = 1. Answer. Joint probability distribution of X and Y is H HH y H x HH 0 1 h(y) 0 1 2 g(x) 0.05 0.25 0.10 0.10 0.30 0.20 0.15 0.55 0.30 0.4 0.6 1 The conditional distribution of Y given X = 1 is w(y|1) = y = 0, 1, 2. That is, that w(0|1) = f (1, y) f (1, y) = for g(1) 0.60 1 f (1, 1) 3 f (1, 2) 2 f (1, 0) = , w(1|1) = = , w(2|1) = = . g(1) 6 g(1) 6 g(1) 6 On the other hand 2 E[Y |1] = 2 y 2 w(y|1) = 02 · y=0 E[Y |1] = 2 yw(y|1) = 0 · y=0 11 1 3 2 + 12 · + 22 · = , 6 6 6 6 3 2 7 1 +1· +2· = . 6 6 6 6 Since the conditional variance of Y given that X = 1 is σY2 |1 = E[Y 2 |1] − (E[Y |1])2 we ﬁnd that σY2 |1 MCB1007 - Int. to Prob. and Statistics 11 = − 6 2 2 17 7 = . 6 36 Second Midterm 17 points Flip a fair coin 3 times. Let X be the number of heads in the ﬁrst 2 ﬂips and let Y be the number of heads on the last 2 ﬂips. Compute Cov(X, Y ). Answer. With 3 tosses there are 8 outcomes in the sample space S = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T }. So, 1 1 f (0, 0) = P (T T T ) = , f (0, 1) = P (T T H) = , f (0, 2) = 0, 8 8 2 1 1 f (1, 0) = P (HT T ) = , f (1, 1) = P (T HT ) = P (HT H) = , f (1, 2) = P (T HH) = , 8 8 8 1 1 f (2, 0) = 0, f (2, 1) = P (HHT ) = , f (2, 2) = P (HHH) = . 8 8 Thus we have the following table HH y H x HHH 0 1 2 h(y) 0 1/8 1/8 0 2/8 1 2 1/8 0 2/8 1/8 1/8 1/8 4/8 2/8 g(x) 2/8 4/8 2/8 1 On the other hand, since μX = E(X) = 2 xg(x) = 0 · g(0) + 1 · g(1) + 2 · g(2) = 0 · 4 2 2 + 1 + 2 · = 1, 8 8 8 yh(y) = 0 · h(0) + 1 · h(1) + 2 · h(2) = 0 · 4 2 2 + 1 + 2 · = 1, 8 8 8 x=0 μY = E(Y ) = 2 y=0 μXY = E(XY ) = 2 2 xyf (x, y) = 1 · 1 · x=0 y=0 1 1 1 10 2 +1·2· +2·1· +2·2· = 8 8 8 8 8 then we have Cov(X, Y ) = σXY = E(XY ) − E(X)E(Y ) = MCB1007 - Int. to Prob. and Statistics 3 10 2 −1·1= . 8 8 Second Midterm 10 + 7 points It is known that 3% of the circuit boards from a production line are defective. If a random sample of 120 circuit boards is taken from this production line (a) determine the probability that the sample contains exactly 2 defective boards. Answer. Substituting x = 2, n = 120, and θ = 0.03 into the formula for the binomial distribution, we obtain 120! 120 × 0.032 × 0.97118 = 0.1766059636. b(2; 120, 0.03) = 0.032 (1 − 0.03)118 = 2! · 118! 2 (b) use the Poisson approximation to estimate the probability that the sample contains 2 defective boards and compare this result with (a). Answer. Substituting x = 2, λ = nθ = 120 × 0.03 = Poisson distribution, we get p(2; 3.6) = 18 5 = 3.6 into the formula for 3.62 × e−3.6 = 0.1770577215. 2! Since n = 120 ≥ 10 and nθ = 3.6 < 10 we see that the approximation obtained in (b) 0.1770577215 is very close the exact probability 0.1766059636. 10 + 5 points A manufacturer received an order of 250 computer chips. Unfortunately, 12 of the chips are defective. To test the shipment, the quality-control engineer randomly selects 20 chips from the box of 250 and tests them. (a) What is the probability of obtaining 3 defective chips? (Leave your answer in terms of factorials) Answer. Substituting x = 3, n = 20, N = 250, and M = 12 into the formula for the hypergeometric distribution, we get 12! 238! 12 238 × = 3! · 9! 17! · 221! . h(3; 20, 250, 12) = 3 25017 250! 20 20! · 230! (b) How many defective chips would you expect to select? Answer. The mean of the hypergeometric distribution is μ = number of defective chips is μ= MCB1007 - Int. to Prob. and Statistics nM . N So the expected 20 · 12 24 nM = = = 0.96. N 250 25 4 Second Midterm 10 + 7 points Suppose that during practice, a basketball player can make a free throw 80% of the time. Furthermore, assume that a sequence of free-throw shooting can be thought of as independent trials. What is the probability that the basketball player makes his (a) tenth free throw on his twelfth shot, (b) ﬁrst free throw on his sixth shot? Answer. (a) Substituting x = 12, k = 10, and θ = 0.8 into the formula for the negative binomial distribution, we get 11 11! 10 2 ∗ b (12; 10, 0.80) = 0.8 0.2 ≈ 0.236. 0.810 (1 − 0.8)2 = 9!2! 9 (b) Substituting x = 6, and θ = 0.8 into the formula for the geometric distribution, we get 4 = 2.56 × 10−4. g(6; 0.8) = 0.8(1 − 0.8)6−1 = 0.8 × 0.25 = 15625 10 + 7 points (a) Find the moment generating function of the random variable whose probability density is given by 2e−2x , for x > 0, f (x) = 0, elsewhere. Answer. The moment generating function of the random variable X is ∞ ∞ c tX tx −2x x(t−2) MX (t) = E[e ] = e 2e dx = 2 e dx = 2 lim ex(t−2) dx c→∞ 0 0 0 c 2 2 2 lim ex(t−2) = lim (ec(t−2) − e0 ) = (0 − 1) = t − 2 c→∞ t − 2 c→∞ t−2 0 2 for t < 2 = 2−t (b) Use the moment generating function to determine μ1 and μ2 for f (x) given in (a). dr MX (t) for r = 1, 2 we obtain Answer. Since μr = dtr t=0 dMX (t) d 1 −1 = (2(2 − t) ) = (2(2 − t)−2 )t=0 = , μ1 = dt dt 2 t=0 t=0 d2 MX (t) d 1 −2 −3 μ2 = (2(2 − t) = ) = (4(2 − t) ) = . t=0 2 dt dt 2 t=0 t=0 MCB1007 - Int. to Prob. and Statistics 5 Second Midterm