STAT 418HW5 SolutionsSections 4.1- 4.31, 5, 7, 13, 17, 20, 21, 25, TE 91.Possible Outcomes 2W 2B 2O 1W1B 1W1O 1B1Ox -2 4 0 1 -1 2P (X = x)0@821A0@1421A0@421A0@1421A10@1421A8·40@1421A8·20@1421A4·20@1421A5. Let X be the difference between the number of heads and number of tails for a cointossed n times. We may observe anywhere from 0 to n heads, where X = −n or X = nrespectively. If we observe n − 1 heads and therefore 1 tail, X = (n − 1) − 1 = n − 2.If we observe n − 1 tails and 1 head, X = 1 − (n − 1) = −(n − 2). If n is even and weobserve the same number of heads and tails, X = 0. If n is even and we observen2+ 1heads andn2− 1 tails, X = 2. If n is even, X = 0, ±2, ±4, . . . , ±(n − 2), ±n or, equivalently,X = n − 2i, i = 0, 1, . . . , n. If n is odd, X = ±1, ±3, . . . , ±(n − 2), ±n or, equivalently,X = n − 2i, i = 0, 1, . . . , n. Therefore X = n − 2i, i = 0, 1, . . . , n.7.a. The wording of this question is unclear. The possible values for the maximum rollbetween the two actual rolls are 1,2,3,4,5 or 6. The maximum roll possible is (6,6).b. Similar to part (a), 1,2,3,4,5, or 6 vs. (1,1).c. 2, 3, . . . , 12d. 0, ±1, ±2, ±3, ±4, ±513. Let X = {total dollar value of all sales}.P (X = 0) = P (no sale on first and no sale on second)= .7 · .4 = .28P (X = 500) = P (one sale and it is for standard)= P (one sale)/2= [P (sale, no sale) + P (no sale, sale)]/2 = [.3 · .4 + .7 · .6]/2 = .27P (X = 1000) = P (2 standard sales) + P (one sale for deluxe)= .3 · .6 ·14+ P (one sale)/2 = .045 + .27 = .315P (X = 1500) = P (2 sales, 1 deluxe, 1 standard)= .3 · .6 ·12= .09P (X = 2000) = P (2 sales, both deluxe)= .3 · .6 ·14= .045117.a.P (X = 1) =12+1 − 14−14=14P (X = 2) =1112−12−2 − 14=16P (X = 3) = 1 −1112=112.b.P12< X <32= F (1.5) − F (.5) =12+1.5 − 14−.54=12.20.a.P (X > 0) = P (W ) + P (LW W ) =1838+203818382= .5918b. No, because if the gambler wins, then he or she wins $1, but a loss would either be $1or $3. E(X) will tell us for sure.c.x 1 -1 -3P (X = x) .5918 P(LWL) + P(LLW) = 21838(2038)2P(LLL) = (2038)3E(X) = .5918 − 2 · .1312 − 3 · .1458 = −.108.21.a. E(X), because whereas the bus driver selected is equally likely to be from any of the4 buses, the student selected is more likely to have come from a bus carrying a largenumber of students.b.E(X) = [402+ 332+ 252+ 502]/148 = 39.28.E(Y ) = [40 + 33 + 25 + 50]/4 = 37.25.E(X) = 60 ·120·320·120+ 20 ·2202·320+ 18 ·2202·120+ 14 ·420·120·620+ 10 ·320·720·620+ 8 ·320·720·120+ 2 ·7202+ 0 ·720·1320·2020− 1 −1202·320−2202·320−2202·120−420·120·620−320·720·620−320·720·120−720·1320·2020−7202!= −.099252TE 9.E(Y ) = E(X/σ − µ/σ) =1σE(X) − µ/σ = µ/σ − µ/σ = 0Var(Y ) = (1/σ)2Var(X) = σ2/σ2=