First Example Exam Solution

First Example Exam Solution - IMSE 213 Prob and Stat First...

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Unformatted text preview: IMSE 213: Prob and Stat First Exam Name: ____________________ Practice First Exam for IENG 213 You may use your text and any "cheat sheets" you created. You may not use your HW papers or the statistical functions on a calculator You may not share texts, cheat sheets, or calculators. 97 Show your work, especially equations used, limits for summations and intergrations, etc. Pts. No. Problem statement Comment: This is based on one of my first tests. 1 What is the probability of drawing 3 aces in a row from a deck of 52 if you: 2 a) b) Replace the card drawn and re-shuffle the deck each time = (4/52)^3 (4/52)^3 comment: too easy too easy 2 Discard each card as it is drawn and re-shuffle the rest of the deck before drawing again =(4/52)(3/51)(2/50) 1 2 A card is drawn from a ordinary 52-card deck. What is the probability that it is a 3, given that it is a face card? ________ , =0 comment: too easy 3 3 Given the following data, compute the median: Data: -4 1 Median: ___________ 2 5 comment: too easy 1.500 =(two middle)/2 4 What is the mean and the standard deviation of the data above? is the mean and the standard deviation of the data above? 2 3 Mean: ___________ Std. Dev.: ___________ 1.000 3.742 comment: too easy 3 3 5 Given that P(A) = 0.6 and P(B) = a) a) P(A U B) = ______ B) 0.680 P(A ∩ B) = ______ b) 0.120 0.2 and A and B are independent. comment: should be word problem 6 6 Three computer brands are each sold with 5 different RAM configurations and 2 different harddisk sizes. How many choices do I have? 30 comment: too easy (c) 2002 Steven Guffey, PhD page 1 9/29/2007 IMSE 213: Prob and Stat First Exam Name: ____________________ 7 7 Find the number of ways 5 can drive more than one car. di th men can be assigned to drive comment: too easy tt N= 5! 4!(5-4)! 4 cars if no driver if indistinquishable cars, then: = 5! 4! 120 = 5 if di ti distinquishable cars, then: i h bl th N= 5! = (5-4)! 5! = 7 8 Find P(A|B) given the following: P(A)= 0.10 P(B)= 0.20 P(B|A) = P(A∩B) P(A) Thus, Thus, P(A|B) = Comment: Replace with total probability problem P(B|A) = P(B|A') = 0.80 0.06 P(A ∩B)= 0.80 (A B) 0.80 P(A∩B) = P(B) 0.08 0.20 = * 0.40 0.10 0.10 = 0.08 0.08 Or: P(A|B) P(A|B) = P( | ) ( ) = (B|A)*P(A) P(B) 0.08 0.20 = 0.40 15 9 For the following continuous frequency distribution, find P(0< x < 0.4) = ________ 3x 2 f ( x) = 0 0 < x <1 elsewhere 0.4 0.4 2 3 P (0 < x < 0.4 = ∫ 3x dx = x 0 = 0 0.064 - 0.000 = 0.064 (c) 2002 Steven Guffey, PhD page 2 9/29/2007 IMSE 213: Prob and Stat First Exam Name: ____________________ 10 For the function to the right: f ( x, y ) = 4 xy , 0 ≤ x ≤1, 0 ≤ y ≤ 1 =0 , otherwise 7 a) Find the marginal distribution of x ∞ 1 1 = 2x − 0 = 2x 0 g ( x) = 7 −∞ ∫ f ( x, y )dy = 0 + ∫ 4 xy dy = 2 xy 2 0 b) Find the marginal distribution y ∞ 1 1 = 2y − 0 = 2y 0 h( y ) = 7 −∞ ∫ f ( x, y )dx = 0 + ∫ 4 xy dx = 2 x 2 y 0 c) Find the conditional probability distribution of x given y : f ( x | y) = f ( x, y ) 4 xy = = 2x h( y ) 2y 0<x<1, 0<y<1 ,y 7 d) Find the probability X is between 0 and 0.1 0.1 P (0 < x < 0.1) = ∫ g ( x ) dx = ∫ 2 x dx = x 0 0 0.1 0.1 2 = 0.01 0 e) Find the probability X <1 100% (c) 2002 Steven Guffey, PhD page 3 9/29/2007 IMSE 213: Prob and Stat First Exam Name: ____________________ e) Determine whether f(x,y) is a possible distribution defined over all possible values sum of P =1 all P>0 ∞ 11 P ( range) = 1 = 1 −∞ ∫ f ( x, y )dxdy = 0 + ∫ ∫ 4 xy dx dy 00 1 2 = ∫ 2 x y dy 0 = ∫ 2 y dy = y 0 0 1 1 2 =1 0 f) Are X and Y independent? Is f(x,y) = g(x) h(y) for all x and y? f(x g(x) h(y) for all and y? 4xy = 2x 2y = 4xy yes 11 Given the joint probability distribution below: f(x,y) y x 1 2 3 g(x) 1 0.10 0.05 0.00 0.15 2 0.05 0.10 0.20 0.35 3 0.05 0.35 0.10 0.50 h(y) 0.20 0.50 0.30 g(x) * h(y) 0.03 0.07 0.03 0.08 0.18 0.18 0.05 0.11 0.05 5 a) b) c) List the marginal distribution of X see above List the marginal distribution of Y (see above) see above Find P(Y=2| X=3) = P(Y=2| =3) f(3 f(3,2) = 0.35 g(3) 0.50 = 0.7 5 5 d) Are X and Y independent ? No, since f(x,y) <> g(x) h(y) for all x and y (c) 2002 Steven Guffey, PhD page 4 9/29/2007 IMSE 213: Prob and Stat First Exam Name: ____________________ More typical questions 0 1 If a there are 40 good and 10 defective fuses, and fuses are selected, what is the probability that all of them will be good? 3 Not independent since choices reduced by previous outcomes: P = 40/50 x 39/49 x 38/48 = 0 0.800 0.796 0.792 = 0.504 2 Given the following table listing the ethnicity of a college class, fill in the frequency : distribution table to the right of the data. No. Ethnic 3 B 4 H 1 AI 12 C 20 Total Frequency Distribution x B H AI C f(x) 0.15 0.2 0.05 0.60 0 3 Is there anything that appears wrong about the following distribution? y g pp g g x f(x) 0 0.28 1 0.22 2 0.33 3 0.17 4 0.11 Answer: 1.11 Exceeds unity (c) 2002 Steven Guffey, PhD page 5 9/29/2007 IMSE 213: Prob and Stat First Exam Name: ____________________ 20 4 For the continuous random variable x, where: a) f(x) = (2 - 3x2), 0 0<x<1 all else Find the mean of the continuous random variable x E(x) = 3x 4 2 2 ∫ x f ( x ) dx = 0 + ∫ x (2 − 3x ) dx = x − 4 −∞ 0 ∞ 1 1 = 1/ 4 0 b) Find the variance of the continuous random variable x [Set up, only] σx2 = E(X2) - μx2 μx = 1/4 1 E(X ) = 2 −∞ ∫ ∞ x 2 f ( x ) dx = 0 + ∫ x 2 (2 − 3x 2 ) dx 0 c) Find the probability that 0 < x < 0.2: 0.2 0.2 2 3 P(0 < x < 0.2) = ∫ (2 − 3x ) dx = 2 x − x 0 = 0.4 − 0.203 = 0 0.39 (c) 2002 Steven Guffey, PhD page 6 9/29/2007 IMSE 213: Prob and Stat First Exam Name: ____________________ 5 Given the following, find the mean of Z and the variance of Z z = 2x - y + 4 μx = 1 μy = 2 σx2 = 1 σy2 = 2 σxy = -3 b -1 μz = = = σz2 = = = a 2 μx 1 + + μx 2 + + c 4 4 a2 σx2 41 18 + + b2 1 σy2 + 2+ 2 2 a 2 b -1 σxy −3 (c) 2002 Steven Guffey, PhD page 7 9/29/2007 ...
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This note was uploaded on 04/03/2009 for the course IENG 213 taught by Professor Staff during the Spring '08 term at WVU.

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