Assignment 2 T109 SolutionsQuestion 1 5 MarksThe quality control manager of Marilyns Cookies is inspecting a batch of chocolate-chip cookies that has just been baked. If the production process is in control, the mean number of chip parts per cookie is .!. "hat is the probability that in any particular cookie being inspected#a$ %ess than fi&e chip parts 'ill be found( 1 mark)ns'er* +#,-.$ / +#,/!$0 +#,/1$ 0 +#,/2$ 0 +#,/3$ 0 +#,/4$ 5rom +oisson table #for mean $, +#,-.$ / !.!!2. 0 !.!146 0!.!44 0 !.!762 0!.1336 = 0.2851#b$ 89actly fi&e chip parts 'ill be found( 1 mark)ns'er* +#, / .$ / 0.1606 #from +oisson distribution table$)lternati&ely, +#, / .$ / .

.:e/ !.1!23#c$ 5i&e or more chip parts 'ill be found( 1 mark)ns'er* + #,;.$ / 1 < +#,-.$ / 1 < !.27.1 / 0.7149#d$ 8ither four or fi&e chip parts 'ill be found( 2 marks)ns'er* +#, / 4 or .$ / +#,/4$ 0 +#,/.$ / !.1336 0 !.1! / 0.2945Question 2 5 Marks)n on-the-job injury occurs once e&ery 1! days on a&erage at an automobile plant. "hat is the probabilitythat the ne9t on-the-job injury 'ill occur 'ithin#a$ 1! days( 2 marks)ns'er* This is a problem in e9ponential distribution, 'here = / !.1! per day and X / 1! days.321 . ! $> 1! # 1! . ! e9p? 1 $ days 1! injury @e9t# = = < P. Therefore, the probability is 0.6321 of the ne9t on-the-job injury to occur 'ithin 1! days.#b$ . days( 2 marks)ns'er* P#@e9t injury - . days$ / 1 < e9p?-!.1!#.$> = 0.3935#c$ 1 day( 1 mark)ns'er* P#@e9t injury - 1 day$ / 1 < e9p?-!.1!#1$> = 0.0952Question 3 5 MarksAur&eys ha&e been conducted on politicians around the 'orld as a 'ay of monitoring the opinions of the electorate. Ai9 month ago, a sur&ey 'as undertaken to determine the degree of support for a @ational +arty politician. Bf a sample of 1!!!, .C indicated that they 'ould &ote for the politician. This month, another sur&ey of 7!! &oters re&ealed that 4C no' support the politician.#a$)t the .C significance le&el, can 'e infer that the politicians population has decreased( 2 marks)ns'er* %o'er confidence limit for the earlier sur&ey is /np pz p$ 1 # /1!!!$ . . ! 1 # . . !6 . 1 . . ! /!..262. Theupperconfidencelimit forthelatersur&eyis/np pz p$ 1 # + / 7!!$ 4 . ! 1 # 4 . !6 . 1 4 . !+ / !.464.. Aince the upper limit of the later islo'er than the lo'er limit of the former #the confidence inter&als do not o&erlap$ 'e can statethat the proportion of the population 'ho support of the politician has signifiantl! "eline" at5# signifiane le$el. #@ote that there are other 'ays of sol&ing this problem.$#b$ )t the .C significance le&el, can 'e infer that the politicians population has decreased bymore than .C( 2 marks)ns'er* The difference bet'een the t'o probabilities as estimated in part #a$ is !..262 < !.464./!.!34D. The difference in proportion 3.4DC is less than .C, and therefore,%e annot inferthat the &olitiian's su&&ort &o&ulation has "erease" (! more than 5#.#c$ 8stimate the decrease in percentage support bet'een no' and si9 months ago.1 mark)ns'er* The e)&ete" $alue of the "erease in &erentage su&&ort is 10 &erent #/!.. < !.4$, but it is at least 3.47# at 5# le$el of signifiane.Question 4 5 Marks) market researcher for a consumer electronics company 'ants to study the tele&ision &ie'ing habits of residents of a particular area. ) random sample of 4! respondents is selected, and each respondentis instructed to keep a detailed record of all tele&ision &ie'ing in a particular 'eek. The results are as follo's* Eie'ing time per 'eek*3 . 1. = X hours, A / 3.7 hours. 2D respondents 'atch the e&ening ne's on at least 3 'eeknights.(a) Construct a 6.C confidence inter&al estimate for the mean amount of tele&ision 'atched per 'eek in this city. 1 *ark)ns'er* Confidence inter&al / nSt Xn 1 , but since the sample siFe is greater than 3! 'e can also use the e9pression nSZ X . The t &alue 'ith 36 degrees of freedom is 2.!2 and the G &alue for 6.C confidence inter&al is 1.6. The confidence inter&al is +14.09, 16.51- using t &alue and +14.12, 16.48- using normal distribution.(b) Construct a 6.C confidence inter&al estimate for the population proportion 'ho 'atch the e&ening ne's on at least 3 'eeknights per 'eek.1 *ark)ns'er* p / 2DH4! / !.D.. Confidence inter&al / np pz p$ 1 # /4!$ D. . ! 1 # D. . !6 . 1 D. . ! / +0.5298, 0.8202-Auppose that the market researcher 'ants to take another sur&ey in a different city. )ns'er these questions*(c) "hat sample siFe is required to be 6.C confident of estimating the population mean to 'ithin 2hours and assumes that the population standard de&iation is equal to . hours(1 *ark)ns'er* Aample siFe, 22 2eZn= / 22 22. $ 6 . 1 #/ 24.!1, 'hich should be rounded up to 25.(d) "hat sample siFe is needed to be 6.C confident of being 'ith in !.!3. of the population proportion 'ho 'atch the e&ening ne's on at least 3 'eeknights if no pre&ious estimate is a&ailable(1 *ark)ns'er* Aample siFe, 22$ 1 #ep p Zn= /22$ !3. . ! #$ . . ! 1 # . . ! $ 6 . 1 # / 784, 'hich is the ma9imum siFe because no pre&ious estimate is a&ailable and p/!.. gi&es the ma9imum siFe.(e) Iased on #c$ and #d$, 'hat sample siFe should the market researcher select if a single sur&ey is being conducted( 1 *ark)ns'er* To meet both the criteria the sample siFe of 784 should be selected in a single sur&ey.