Introduction to Statistics Homework 4 Solutions

Math 130 Solutions – Introduction to Statistics Homework 4 Solutions Assignment Chapter 15: 2, 6, 20, 31, 43 Chapter 16: 9, 20, 24, 28, 47 Hint on 16.47] Think about the mean and standard deviation of a combination of six random variables, one for each day the shop is open. Chapter 15 15.2] Travel. The given information: Let M = event that a U.S. resident has traveled to Mexico Let C = event that a U.S. resident has traveled to Canada P(M) = 0.09 P(C) = 0.18 M C 0.05 0.04 0.14 P(M and C) = 0.04 0.77 A Venn diagram is useful here. (a) P(C but not M) = 0.14. There’s a 14% chance that a US resident has traveled to Canada but not Mexico. (b) P(C or M) = P(C) + P(M) – P(C and M) = 0.18 + 0.09 – 0.04 = 0.23. Or, add the circular sections of the Venn diagram to get 0.05 + 0.04 + 0.14 = 0.23. There’s a 23% chance that a U.S. resident has traveled to Canada or Mexico. (c) P(neither C or M) = 1 – P(C or M) = 1 – 0.23 = 0.77 There’s a 77% chance that a randomly selected resident has traveled to neither Canada nor Mexico. 15.6] Birth Order. (a) (b) (c) (d) Human Ecology = Ecology is 0.1928. irst-born" = ## = 0.1928. The probability that a student is in Human = 0.5067. The probability that a student is a first born is 0.5067. irst-born and Ecology" = = 0.0673. The probability that a student is a firstborn and Ecology student is 0.0673. irst-born or Ecology" #( = irst-born" + Ecology − = 0.5067 + 0.1928 − 0.0673 = 0.6322 irst-born and Ecology" The probability that a student is a first-born or Ecology student is 0.6322. 15.20] Benefits. A Venn diagram can help us here. Let R = Retirement plan Let H = Health Insurance (a) P(not R and not H) = 0.25 Using probability rules: not R and not H = 1 − R or H = 1 − [ . + / + R and H ] = 1 − [0.56 + 0.68 − 0.49] = 0.25 There is a 25% probability that he has neither a retirement plan nor employer sponsored health insurance. (b) /|. = 3 H and R 3 4 = 5. 6 5.(7 retirement plan is 0.875. ? = 0.875. The probability of health insurance given he has a (c) No, they are not independent, because / and . = 8 ? / 0.49 = 8 0.68 0.56 0.49 ≠ 0.3808 . (d) They are not mutually exclusive, because there is some overlap between the two events. There are cases with both R and H. 15.31] Montana. Party affiliation is not independent of sex. To see this, let D represent Democrats, and M represent males. < 3 > and ? 7 Now ; = = 0.4158 and ;|= = = = 0.3429 Since 5 ;|= ≠ 3 ? ; , these events are not independent. #5( 15.43] Dishwashers. A tree diagram can help us here. Chuck|Break) = = 3(Chuck and Break) 3(Break) (5. )(5.5#) (5. )(5.5#)F(5. )(5.5#)F(5. )(5.5 ) 5.556 = 5.55 5.556 F5.55 F5.556 = 5.5#7 = 0.5625 If we hear a dish break, there’s a 56.25% chance that Chuck is working. Chapter 16 16.9] Software. The given information is: Large Contract: $50,000 profit Small Contract: $20,000 profit 30% chance 60% chance Both Contracts: $70,000 profit (0.30)(0.60) = 18% chance Let X be the profit. The distribution of X can be tabulated as X P(X) 50,000 0.12 20,000 0.42 70,000 0.18 0 0.28 K G -H0 = I HJ (HJ ) = 50000(0.12) + 20000(0.42) + 70000(0.18) + 0(0.28) = $27,000 JL# We’d expect an average profit of $27,000 . 16.20] Insurance. We have the following information: Costs $100 Major Injury: Pays $10,000, Probability of 1/2000 Minor Injury: Pays $3,000, Probability of 1/500 Let X be the profit. (a) We can make a table for the probability model. X P(X) 100 0.9975 -9,900 0.0005 -2900 0.002 K (b) The company’s expected profit is $89. G -H0 = I HJ (HJ ) = 100(0.9975) − 9900(0.0005) − 2900(0.002) = $89. JL# (c) The standard deviation is $260.54. N K = OI(HJ − P) JL# (HJ ) = Q(100 − 89) (0.9975) + (−9900 − 89) (0.0005) + (−2900 − 89) (0.002) = √120.6975 + 49890.0605 + 17868.242 = √67879 = 260.536 16.24] Contracts. We have the following information: • • • • We bid on two contracts P(contract #1) = 0.8 P(contract #2) = 0.2, if you get contract #1 P(contract #2) = 0.3, if you don’t get contract #1 (a) No, the two contracts are not independent. The probability of the second contract changes depending on whether or not we got the first contract. The two events are dependent. (b) There are several approaches that might work here. Let A = the event that we get contract #1 Let B = the event that we get contract #2 We want to find P(A and B). We can use the general multiplication rule here. S and T = S T|S = 0.8 0.2 = 0.16 There’s a 16% probability that we get both contracts. We could also try a tree diagram: Contract #1 Contract#2 0.2 Yes (0.8)(0.2) = 0.16 0.8 No (0.8)(0.8) = 0.64 0.3 Yes (0.2)(0.3) = 0.06 0.7 No (0.2)(0.7) = 0.14 Yes 0.8 0.2 No (c) Let AC = the event that we don’t get contract #1 Let BC = the event that we don’t get contract #2 We want to find P(AC and BC). We can use the general multiplication rule again. SU and T U = SU T U |SU = 0.2 0.7 = 0.14 There’s a 6% probability that we get no contracts. (d) We can make a table for the probability model. X P(X) (e) We have: N K G [H ] = I HJ K JL# = OI HJ − P JL# = Q 0 − 1.02 0 0.14 1 0.70 2 0.16 HJ = 0 0.14 + 1 0.70 + 2 0.16 = 1.02 HJ 0.14 + 1 − 1.02 = √0.1457 + 0.0028 + 0.1537 = √0.3022 = 0.5497 0.70 + 2 − 1.02 0.16 16.28] Random Variables. Find the mean and standard deviation of each of these variables. (a) X – 20 Mean = G -H − 200 = G -H0 − 20 = 10 − 20 = −10 Var(H − 20) = Var(H) = 4 ⇒ Y; = 2 (b) 0.5Y Mean = G -0.5Z0 = 0.5G -Z0 = 0.5(20) = 10 Var(0.5Z) = 0.5 Var(Z) = 0.25(25) = 6.25 ⇒ Y; = 2.5 (c) X + Y Mean = G-H + Z0 = G-H0 + G-Z0 = 10 + 20 = 30 Var(H + Z) = Var(H) + Var(Z) = 4 + 25 = 29 ⇒ Y; = 5.39 (Note: The variances add since X and Y are independent) (d) X – Y Mean = G -H − Z0 = G -H0 − G -Z0 = 10 − 20 = −10 Var(H − Z) = Var(H) + Var(Z) = 4 + 25 = 29 ⇒ Y; = 5.39 (e) Y1 + Y2 Mean = G -Z# + Z 0 = G -Z# 0 + G -Z 0 = 20 + 20 = 40 Var(Z# + Z ) = Var(Z# ) + Var(Z ) = 25 + 25 = 50 ⇒ Y; = 7.07 16.47] Coffee and Doughnuts. X Y X – 20 0.5Y X+Y X–Y Y1 + Y2 Mean SD 10 2 20 5 -10 2 10 2.5 30 5.39 -10 5.39 40 7.07