Math 17, Section 2 – Spring 2011 Homework 4 Solutions

Math 17, Section 2 – Spring 2011 Homework 4 Solutions Assignment Chapter 15: 2, 6, 20, 31, 43 Chapter 16: 9, 20, 47 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)
Human Ecology    0.1928. The probability that a student is in Human Ecology is 0.1928. (b)
irst-born"  ##   0.5067. The probability that a student is a first born is 0.5067. #( (c)
irst-born and Ecology"    0.0673. The probability that a student is a firstborn and Ecology student is 0.0673. (d)
irst-born or Ecology" 
irst-born" *
Ecology +
irst-born and Ecology"  0.5067 * 0.1928 + 0.0673  0.6322 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 H0  1 + -0.56 * 0.68 + 0.490  0.25 There is a 25% probability that he has neither a retirement plan nor employer sponsored health insurance. (b)
/|.   3H and R 34  5.6 5.(7  0.875. The probability of health insurance given he has a retirement plan is 0.875. (c) No, they are not independent, because ?
/ and .   8
/ 
. ? 0.49  8 0.680.56 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 5 3? Since
;|= :
;, these events are not independent. #5( 15.43] Dishwashers. A tree diagram can help us here.
Chuck|Break   3Chuck and Break 3Break 5.5.5# 5.5.5#F5.5.5#F5.5.5 5.556  5.55F5.55F5.556 5.556  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   500000.12 * 200000.42 * 700000.18 * 00.28  $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   1000.9975 + 99000.0005 + 29000.002  $89. JL# (c) The standard deviation is $260.54. N K  OIHJ + P
HJ  JL#  100 + 89 0.9975 * +9900 + 89 0.0005 * +2900 + 89 0.002  √120.6975 * 49890.0605 * 17868.242  √67879  260.536 16.47] Coffee and Doughnuts. This problem is not graded, as I haven’t covered that material during our lecture.