Math 151. Rumbos Spring Solutions to Assignment #7
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1 Math. Rumbos Sping 202 Solutions to Assignment #7. Fo each of the following, find the value of the constant c fo which the given function, p(x, is the pobability mass function (pmf of some discete andom vaiable. ( x 2 (a p(x = c, fo x =, 2, 3,... and zeo else. 3 Solution: We find c fo which p(k = ; that is, fo which c It then follows that c = /2. ( k 2 =, 3 ( k 2 = 2/3 3 2/3 = 2. (b p(x = cx fo x =, 2, 3, 4,, and zeo othewise. Solution: We find c fo which p(k = ; that is, fo which It then follows that c = /. c k =, k = 6 2 =.
2 Math. Rumbos Sping Suppose that two balanced dice ae olled, and let X denote the absolute value of the diffeence between the two numbes that appea. Detemine and and sketch the pmf of X. Solution: The possible values fo X ae 0,, 2, 3, 4,. The sample space is shown in the table below (, (, 2 (, 3 (, 4 (, (, 6 (2, (2, 2 (2, 3 (2, 4 (2, (2, 6 (3, (3, 2 (3, 3 (3, 4 (3, (3, 6 (4, (4, 2 (4, 3 (4, 4 (4, (4, 6 (, (, 2 (, 3 (, 4 (, (, 6 (6, (6, 2 (6, 3 (6, 4 (6, (6, 6 the fist enty in the pai (m, n indicates numbe on the fist die, and the second enty that on the second die. Since the dice ae balanced, the outcomes ae equally likely. Thus, all the outcomes have pobability /36. The outcomes that yield X = 0 lie along the main diagonal and thee ae six of them. It the follows that P(X = 0 = 6 36 = 6. To find P(X =, the table below shows those outcomes that yield X = undelined (, (, 2 (, 3 (, 4 (, (, 6 (2, (2, 2 (2, 3 (2, 4 (2, (2, 6 (3, (3, 2 (3, 3 (3, 4 (3, (3, 6 (4, (4, 2 (4, 3 (4, 4 (4, (4, 6 (, (, 2 (, 3 (, 4 (, (, 6 (6, (6, 2 (6, 3 (6, 4 (6, (6, 6 Thee ae 0 of those events. It then follows that P(X = = 0 36 = 8. We poceed in a simila fashion computing the values fo P(X = k fo k = 2, 3,. Fo instance, in the case k = 2, the undelined outcomes would be tw0 steps emoved fom the main diagonal. Thee ae eight of those; consequently, P(X = 2 = 8 36 = 2 9.
3 Math. Rumbos Sping Similaly, and P(X = 3 = 6 36 = 6, P(X = 4 = 4 P(X = = 2 36 = 9, 36 = Suppose that a box contains seven ed balls and thee blue balls. If five of them ae selected at andom, without eplacement, detemine the pmf of the numbe of ed balls that will be obtained. Solution: Let X denote the numbe of ed balls in the five selected ones. Then X is discete and takes on the possible values 2, 3, 4 and. We have that P(X = k (goups of k ed balls out of 7(goups of k blue balls out of 3 = goups of balls out of 0 ( 7 3 = k( k ( 0 fo k = 2, 3, 4,, ( n = k n! k!(n k! ae the binomial coefficients. 4. A civil enginee is studying a left tun lane that is long enough to hold 7 cas. Let X denote the numbe of cas left in the lane at the end of andomly chosen ed light. The enginee believes that the pobability that X = x is popotional to (x + (8 x fo x = 0,,..., 7 (the possible values of X. (a Find the pmf fo X.
4 Math. Rumbos Sping Solution: We have that p X (x = c(x+(8 x fo x = 0,,..., 7 and some constant of popotionality c. To find c, we use the fact that 7 p X (k =, o c k=0 7 (k + (8 k =, k=0 7 (k + (8 k = = 20. k=0 Thus, c = /20 and and p X (x = (x + (8 x 20 p X (x = 0 othewise. (b Find the pobability that X will be at least. Solution: We compute fo x = 0,, 2,..., 7; P(X = P(X = + P(X = 6 + P(X = 7 = = = 3.. Select five cads at andom and without eplacement fom an odinay deck of playing cads. Let X denote the numbe of heats in the five cads. (a Find the pobability mass function (pmf of X. Denote it by p(x.
5 Math. Rumbos Sping 202 Solution: The possible values fo X ae 0,, 2, 3, 4 and. To compute p(x, we poceed as in Poblem 3: p(k = ( 3 ( 39 k k ( 2 fo k = 0,, 2, 3, 4,. (b Detemine P(X. Solution: Compute P(X = p(0 + p(, and p(0 = ( 3 0 p( = ( 39 ( 2 = ( 3 ( 2 ( 39 ( 2 = ( 39 4 = 3 ( 39 4 ( Thus, P(X o 63.29%. (c Find the cumulative distibution function, F (x = P(X x, and sketch its gaph along with that of p(x. Solution: Completing the calculations fo p(x we get if x = 0, if x =, if x = 2, p(x if x = 3, if x = 4, if x =, 0 othewise. The gaphs of p and F X ae shown in Figues and 2, espectively.
6 Math. Rumbos Sping p x Figue : Pobability Mass Function fo X F X x Figue 2: Cumulative Distibution Function fo X
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