Stat 11 Homework 3 Due 5/12/04 1. Of Colossal Conglomerate's 16,000 clients, 3,200 own their own business, 1,600 are "gold class" customers, and 800 own their own business and are also "gold class" customers. What is the probability that a randomly chosen client who owns his or her own business is a "gold class" customer? If B= own their own business and A=gold class answer: .25 solution 3200800)()()|( =∩=BPBAPBAP 2. You have invested in Home-Clone Inc. stocks, as you suspect that the company's "Clone-a-Sibling" kit will shortly be approved by the FDA. There is an 80% chance that FDA approval will be given, and a 95% chance that the value of the stock you hold will double if FDA approval is given. What is the probability that the FDA will approve the product and the value of the stock you hold will double? Let A= stock doubles B=FDA approval is given recall )()()|(BPBAPBAP∩= rearranging terms )()|()( BPBAPBAP=∩ so .80*.95 = .76 or 76% 3. In a certain population, 30% of the persons have accounting training and 8% have a computer training. Moreover, 12% of the persons who have accounting training have computer training. let A= accounting training and B=computer training A. What percent of the population that have accounting training has computer training? Sorry it’s kind of poorly written. Of the entire population, 12% of 30% or 3.6% of the entire population have both. If you think the population is just those with accounting training that’s 12%. The answer I am looking for is 3.6% or .036 B. What percent of the population with computer training also have accounting training? 45.08.)12.*3(.)()()|( ==∩=BPBAPBAP C. Are computer training and accounting training independent? No: )()()()()|()()|(BPAPBAPBPABPAPBAP⋅≠∩≠≠ )024.0036.0()30.012.0()12.045.0(≠≠≠Stat 11 Homework 3 Due 5/12/04 D. Are computer training and accounting training mutually exclusive? No: 036.012.03.0)(≠=⋅=∩ BAP 4. Gondor Stewards, an auditing firm, has been hired to audit the books for the Rivendell Marketing Research. Anduin Took, Gondor Stewards top auditor, has been given this assignment. Rivendell, which has twenty-three clients of their own, only wants a quick overview rather than a complete audit. Because of this, Ms. Took decides to only audit the accounts for five randomly selected Rivendell clients. The results of her audit are provided below: Rivendell Client Number of Correct Entries Number of Incorrect Entries Misty Mountain Furniture Makers 1219 38 Greenleaf’s Appliance Production 2436 112 Rose Cotton’s Maid Service 844 11 Fangorn Dry Cleaners 1856 39 Mordor Kennels 564 17 Before we start answering the proceeding questions, let’s agree on some notation for particular events in our data. We’ll let A = Correct Entry (so A’ = Incorrect Entry), B1 = Misty Mountain Furniture Makers, B2 = Greenleaf’s Appliance Production, B3 = Rose Cotton’s Maid Service, B4 = Fangorn Dry Cleaners, and B5 = Mordor Kennels. A. Using the data collected by Ms. Took, what would you estimate to be the probability of an incorrect entry? 0304.7136217)'( ===totalincorrectAP B. Ms. Took has selected Mordor Kennels at random. What would you estimate to be the conditional probability of an incorrect entry in their books? 0291.7136/)17564(7136/17)5()5'()5|'( =+=∩=BPBAPBAP C. What would you estimate to be the probability that an entry in Rivendell’s books is drawn from a manufacturing organization (Misty Mountain Furniture Makers or Greenleaf’s Appliance Production)? P(B1 ∪ B2) = P(B1)+ P(B2) =5332.7136)1122436(7136)381219(=+++Stat 11 Homework 3 Due 5/12/04 D. What would you estimate to be the probability that an entry in Rivendell’s books is drawn from a service organization (Rose Cotton’s Maid Service or Fangorn Dry Cleaners or Mordor Kennels)? 1.0 – .5332 = .4668 (use the results from the previous question) E. What would you estimate to be the probability that an incorrect entry found in the books is from a service organization? let’s call the service organizations B345 P(B345|A’)= 3088.)7136/217(7136/67)'()'345(==∩APABP F. Restructure the table so that it is simply manufacturing & service organization versus correct and incorrect (we don’t care about the names any more, just the type). Do the type of industry (service vs. manufacturing) and the status of an entry in the books (correct or incorrect) appear to be independent? Please explain. Correct (A) Incorrect (A’) Manufacturing (B12) 3655 150 Service (B345) 3264 67 For this to be true, then for example, if we use the results from the previous problem, P(B345|A’) = P(B345) P(B345|A’)= 3088.)7136/217(7136/67)'()'345(==∩APABP is not the same as P(B345) which is .4574 5. In a certain zip code it is known that 80% of the adult women shop online. If 12 women from this zip code are selected at random, what is the probability that 9 of them will shop online? Please assume independence. [section 4.3, you can’t do this until after 5/5/04 lecture] answer: p(9) = 2362.008.*1342.*1*2*310*11*12)008(.*)1342(.*!3!9!12)20(.)80(.91239===⎟⎟⎠⎞⎜⎜⎝⎛ see page 117-118 6. There are 27 students in a discussion section in Statistics. Let x be the random variable which gives the age (rounded to the nearest year) of a randomly chosen student on the day of the final exam. (i) Is X a discrete or a continuous random variable ? Is x a finite random variable ? Explain. In this context, X is discrete. X is also finite since its range of possible values is limited (probably 17-25, perhaps higher, but it won’t be over 120 or so)Stat 11 Homework 3 Due 5/12/04 (ii) Assume that on the day of the final exam, 3 of the students are 17 years old, 5 are 18 years old, 10 are 19 years old, 6 are 20 years old and 3 are 21 years old. Answer the following questions (you may give your answers as fractions): • The p.d.f. f(x) has f(x) =f(19) = 10/27 • P( x < 22) = 27/27 • P(17 < x < 22) = (5+10+6+3)/27 • The c.d.f. F(x) has F(x)=F(20) = (3+5+10+6)/27 (iii) Based on the information given in (ii), find the expected value of x and the standard deviation of x. Show all your work. E(X)=17*(3/27) + 18*(5/27) + 19*(10/27) + 20*(6/27) + 21*(3/27) = 19.037 or about 19 years using the easy formula (p. 114)