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Course: STAT 134, Fall 2008
School: Berkeley
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4653: Math Elementary Probability: Spring 2007 Homework #6. Problems and Solutions 1. Sec. 4.2: #6: A Geiger counter is recording background radiation at an average rate of one hit per minute. Let T3 be the time in minutes when the third hit occurs after the counter is switched on. Find P (2 T3 4). Solution. The random variable T3 has a gamma distribution with parameter r = 3 and = 1, which gives us a density...

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4653: Math Elementary Probability: Spring 2007 Homework #6. Problems and Solutions 1. Sec. 4.2: #6: A Geiger counter is recording background radiation at an average rate of one hit per minute. Let T3 be the time in minutes when the third hit occurs after the counter is switched on. Find P (2 T3 4). Solution. The random variable T3 has a gamma distribution with parameter r = 3 and = 1, which gives us a density function f (t) = Therefore, P (2 T3 4) = 2 4 2 -t t e r tr-1 r-1 -t 1 2 -t t e = t e . (r) 2 2 -4 dt = -e-t t2 +t+1 2 4 2 = -e (8 + 4 + 1) + e-2 (2 + 2 + 1) = 5e-2 - 13e-4 0.4386. 2. Sec. 4.4: #4: Suppose X has uniform distribution on (-1, 1). Find the density of Y = X 2 . Solution. It is easy to show that |X| has uniform distribution on [0, 1). Then Y has density fY (y) = f|X| |dy/d|x| | = 1 2|x| = 1 2 y for 0 < y < 1, otherwise. 0 3. Sec. 4.5: #2: Find the cumulative distribution functions of: a) the binomial (3, 1/2) distribution; b) the geometric (1/2) distribution on {1, 2, . . .}. Solution. 0 for x < 0, 1/8 for 0 x < 1, a) P (X x) = 1/2 for 1 x < 2, 7/8 for 2 x < 3, 1 for x 3. b) P (X x) = 0 2k -1 2k for x < 1, for k x < k + 1, k = 1, 2, . . . . 4. Sec. 4.5: #4: Let X be a random variable with c.d.f. F (x). Find the c.d.f. of aX + b first for a > 0, then for a < 0. Solution. Let Y = aX + b. Then the c.d.f. for Y is FY (y) = P (Y y) = P (aX + b y) + P (aX y - b). For a > 0, we have FY (y) = P X For a < 0, we have FY (y) = P X y-b a =1-P X < 1 y-b a =1-F y-b - . a y-b a =F y-b . a 5. Sec. 4.5: #6: Let X be a random variable with c.d.f. F (x) = x3 for 0 x 1. Find: 1 a) P (X 2 ); b) the density function f (x); c) E(X). d) Let Y1 , Y2 , Y3 be three points chosen independently and uniformly on the unit interval, and let X be the rightmost point. Show that X has the distribution described above. Solution. a) P (X 1/2) = 1 - F (1/2) = 1 - (1/2)3 = 7/8. b) Since f (x) = F (x), we have f (x) = c) 3x2 0 for 0 x 1, for x < 0 or x > 1. 1 E(X) = - xf (x)dx = - 3 x 3x2 dx = x4 4 0 3 = . 4 d) Let Y be a random variable uniformly distributed on the unit interval (0, 1), let FY (y) denote its c.d.f., and let FX (x) denote the c.d.f. for X = max(Y1 , Y2 , Y3 ). Then 0 for x < 0, 3 3 for 0 x 1, FX (x) = [FY (x)] = x 1 for x > 1, which is the c.d.f. described above. 6. Sec. 4.6: #2: Let X have beta (r, s) distribution. a) Find E(X 2 ), and use the formula for E(X) given in this section to find Var (X). b) Find a formula for E(X k ), for integers k 1. Solution. a) We have 1 1 B(r + 2, s) xr+1 (1 - x)s-1 dx = B(r, s) 0 B(r, s) (r + 2)(s) (r + s) r(r + 1) = = , (r + s + 2) (r)(s) (r + s)(r + s + 1) r2 rs r(r + 1) Var (X) = E(X 2 ) - [E(X)]2 = - = . 2 2 (r + s + 1) (r + s)(r + s + 1) (r + s) (r + s) E(X 2 ) = b) We have E(X k ) = 1 1 B(r + k, s) xr+k-1 (1 - x)s-1 dx = B(r, 0 s) B(r, s) (r + k)(s) (r + s) r(r + 1) (r + k - 1) = = . (r + s + k) (r)(s) (r + s)(r + s + 1) (r + s + k - 10 7. Sec. 5.1: #2: A metal rod is l inches long. Measurements on the length of this rod are equal to l plus a random error. Assume that the errors are uniformly distributed over the range -0.1 inch to +0.1 inch, and are independent of each other. a) Find the chance that a measurement is less than 1/100 of an inch away from l. b) Find the chance that two measurements are less than 1/100 of an inch away from each other. 2 Solution. a) Let X be the measurement error. Then P (|X| < 0.01) = 0.02 = 0.1. 0.2 b) Let Y be the second measurement error, and let the set of possible values of (X, Y ) be S = {(x, y) : |x| < 0.1 and |y| < 0.1}. Then the region where the two measurement are less than 1/100 of an inch away from each other is B = {(x, y) S : |x - y| < 0.01}. Then P (|X - Y | < 0.01) = area (B) 0.22 - 0.192 39 = = = 0.0975. area (S) 0.22 400 8. Sec. 5.1: #6a): A group of 10 people agree to meet for lunch at a cafe between 12 noon and 12 : 15 P.M. Assume that each person arrives at the cafe at a time uniformly distributed between noon and 12 : 15 P.M., and that the arrival times are independent of each other. Jack and Jill are two members of the group. Find the probability that Jack arrives at least two minutes before Jill. Solution. Let X be Jacks arrival time, measured in minutes after 12 noon, and let Y be Jills arrival time, measured on the same scale. Then P (X < Y - 2) = 169 132 /2 = 0.3756. 152 450 9. Sec.5.2: #4: For random variables X and Y with joint density function f (x, y) = 6 e-2x-3y (x, y > 0) and f (x, y) = 0 otherwise, find a) P (X x, Y y); b) fX (x); c) fY (y). d) Are X and Y independent? Give a reason for your answer. Solution. a) y x y x P (X x, Y y) = - - y f (t, s) dtds = 0 x 0 y 6e-2t e-3s dtds y x 0 = 0 3e -3s 0 2e y -2t dtds = 0 3e -3s -e -2t ds = 0 3e-3s (1 - e-2x )ds = (1 - e-2x ) 0 3e-3s ds = (1 - e-2x )(1 - e-3y ). b) fX (x) = - f (x, y)dy = 0 6e -2x -3y e dy = 2e -2x 0 3e-3y dy = 2e-2x . c) fY (y) = - f (x, y)dx = 0 6e-2x e-3y dx = 3e-3y 0 2e-2x dx = 3e-3y . d) X and Y are independent because f (x, y) = fX (x)fY (y). 3 10. Sec. 5.2: #8a): Random variables X and Y have joint density fX,Y (x, y) = c(y 2 - x2 )e-y 0 for - y x y, y > 0; otherwise. Here c is a constant. Show that Y has a gamma density, and hence deduce that c = 1/8. Solution. For y > 0, the marginal density for Y is y fY (y) = - fX,Y (x, y)dx = -y c(y 2 - x2 )e-y dx = ce-y y 2 x - (-y)3 3 4c 3 -y y e . 3 x3 3 x=y x=-y ce-y y3 - y3 3 - (-y)3 - = Recall that, for y > 0, the density function for the gamma distribution with parameters r and is fr, (y) = r y r-1 -y e . (r) We see that Y has gamma distribution with parameters r = 4 and = 1 and therefore 1 1 4c = = , 3 (4) 3! 1 c= . 8 4
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Berkeley - STAT - 134
Math 4653: Elementary Probability: Spring 2007Homework #4. Problems and Solutions1. Sec. 3.1: #8a): A hand of five cards contains two aces and three kings. The five cards are shuffled and dealt one by one, until an ace appears. Display in a table the di
Berkeley - STAT - 134
Math 4653: Elementary Probability: Spring 2007Homework #2. Problems and Solutions (corrected)1. Sec. 1.5: #2: Polyas urn scheme. An urn contains 4 white balls and 6 black balls. A ball is chosen at random, and its color noted. The ball is then replaced,
Berkeley - STAT - 134
Berkeley - STAT - 134
Math 4653: Elementary Probability: Spring 2007Homework #1. Problems and Solutions1. Appendix 1 (vi): Prove that 2n nn=k=0n kn n-kn=k=0n k2.Solution. The left side is the number of all subsets of the set cfw_1, 2, . . . , n-1, n, n+1, . . . ,
Berkeley - STAT - 134
Math 4653: Elementary Probability: Spring 2007Homework #3. Problems and Solutions1. Sec. 2.4: #2: Find Poisson approximations to the probabilities of the following events in 500 independent trials with probability 0.02 of success on each trial: a) 1 suc
Berkeley - STAT - 134
Math 4653: Elementary Probability: Spring 2007Homework #5. Problems and Solutions1. Sec. 3.5: #2: How many raisins must cookies contain on average for the chance of a cookie containing at least one raisin to be at least 99%? Solution. Let X be the numbe
Berkeley - STAT - 134
Math 4653: Elementary Probability: Spring 2007Homework #7. Problems and Solutions1. Ch. 4, Review: #21: Suppose R1 and R2 are two independent random variables with the 1 same density function f (x) = x exp(- 2 x2 ) for x 0. Find a) the density of Y = mi
Berkeley - STAT - 134
STAT516 Solution to Homework 3Section 2.1 4 1 6 1 = 0.375 2. P (2 boys &amp; 2 girls) = = 2 2 2 16 Hence, P (different number of boys &amp; girls in a family of 4 children) = 1 - P (2 boys &amp; 2 girls) = 1 - 0.375 = 0.625 So, in a family of 4 children, different n
Berkeley - STAT - 134
STAT516Section 3.2FALL 2005Solution to Homework 52. a) average(3rd list) = average(1st list) + average(2nd list) = 5.8 b) average(3rd list) = average(1st list) average(2nd list) = -2.2 c) &amp; d) Can't do it: need to know the order of the numbers in the
Berkeley - STAT - 134
Solution to Homework 8Section 4.2 1. Let X denote the lifetime of an atom. Then X has an exponential distribution with = log 2 (as half life is 1). a) P ( X &gt; 5) = e -5 = 1/ 32 . b) P ( X &gt; t ) = .1 e - t = .1 t = (log10) / t = 3.32 . c) Assume that life
Berkeley - STAT - 134
Section 5.1 1. a) 7/124b) 5/36.411 1 200122. a) 0.1 b) 1 - 2(1/ 2)(0.19) 2 = 0.0975 (0.2) 20.20.01 0.014. a) 1 - ( 3 / 4 ) = 7 /16 = 0.437520.210.250 0.25 1X b) P - 1 Y 4 0.25 P X = 5 Y4 1 4 3 X 1 - + 9 / 40 = 0.225 = = 3 2 5 43 4
Berkeley - STAT - 134
Section 5.4 1. a)r , 12X2 1r1, 11 dx if 0 &lt; z &lt; 1 2 0z0b) f X1 + X 2 ( z ) =f X1 ( x) f X 2 ( z - x)dx=1 1 x&lt;2 2 dx = 0&lt; x&lt;1;zx&lt; z 2 dx = 0 &lt; x &lt;1;0 &lt; z - - 2&lt;dx 20 111if 1 &lt; z &lt; 21 dx if 2 &lt; z &lt; 3 2 z-2 Xz if 0 &lt; z &lt; 1 2 1 = if 1 &lt; z &lt; 2
Berkeley - STAT - 134
STAT516Section 2.4 1.FALL 2005SOLUTION TO HOMEWORK 4b) See text for Poisson(2) histogram. c) See text and &quot;approximate&quot; Poisson(.3284) histogram.2. a) The number of successes in 500 independent trials with success probability 0.02 has Binomial(500,0.
Berkeley - STAT - 134
Solution to Homework 7Section 4.1 1. Treat the density function as constant over those small intervals. 1 a) (0.001)( )=0.000399 2 1 b) (0.001)( )( e -0.5 )=0.00025 2-4 2. a) cx dx = 1 1 1 cc = 3 b) mean = 3/2 c) variance = 3. a) cx(1 - x)dx = 10 1/
Berkeley - STAT - 134
Section 4.5 6. a) 1 - 1/8 = 7/8. f 3x 2 f ( x) = b) 0 c) E ( X ) = 3 / 4. d) P( X Section 4.6 1. Let X i denote the gap (in minutes) between the ith person's arrival time and noon. Then X i 's are i.i.d. N( 0, 52). a) P (1st person arrives before 11:50) =
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STAT516 Solution to Homework 1 1.1.4: We have 18 black, 18 red, 2 green, total 38 a) P(both lose)=2/38 b) P(at least one wins)=1-P(both lose)=36/38 c) P(at least one loses)=1-P(both win)=1-0=1 1.3.2: a)(AB c ) (Ac B) or A B - AB b)(A B C)c or Ac B c C c1
Berkeley - STAT - 134
Section 3.3 19. Let X j be the weight (in lbs) of the j -th person, j = 1, 2, ., 30. Hence, total weight is: S = X 1 + X 2 + . + X 30 As X j 's are independent with E ( X j ) = 150 and SD( X j ) = 55 , we can use CLT to calculate: 5000 - 30 150 P ( S &gt;\$ 5
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