Question

Complete: Chapter 7 Problem Set Back to Assignment Average: 2/6 Attempts: As Aa 6. Three different distributions There are 559 full-service restaurants in Delaware. The mean number of seats per restaurant is 99.2. [Source: Data based on the 2002 Economic Census from U.S. Census Bureau. Suppose that the true population mean p 99.2 and standard deviation a 20.2 are unknown to the Delaware tourism board. They select a simple random sample of 50 full-service restaurants located within the state to estimate p. The mean number of seats per restaurant in the sample is M- 103.4, with a sample standard deviation ofs 18.2. (Note: The standard deviation of the distribution of sample means (that is, the standard error, on) is Although u and a are unknown to the Delaware tourism board, they are known to you for the purposes of calculating these answers.) The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (u 99.2) and one randomly selected full-service restaurant in Delaware is The standard or typical aveesge difference between the mean number of seats in the sample of 50 restaurants (M 103.4) and one randomly selected restaurant in that sample is The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (u 99.2) and the sample mean of any sample of size 50 is The z-score that locates the mean number of seats in the Delaware tourism board's somple (M- 103.4) in the distribution of sample means is use the unit normal tables and accompanying figures to answer the question that follows. To use the tables, click o the Unit Nonmal Tables tab below the figures, and use the dropdown box to select thie desired range of z-score vatues A table of the proportions of the nomal distribution connesponding to that range of z-scores will appear, If vou need a dferent range of z-scores, simply click on the boX gain and select a naw range suggestion: Hake a sketch of the area under the inormal distribution you ard seking this skatch l hele you deteine wNCN column(s) of thid nomal tebie to use in detemining the approprate pnob aotityi D en Complete: Chapter 7 Problem Set The standard or typical average differene bebween the mean number of seats in the 559 full-service restaurants in Delaware (u-99.2) and one randomly selected full-service restaurant in Delaware is The standard or typical average difference between the mean number of seats in the sample of 103.4) and one randomly selected restaurant In that sample is 50 restaurants (M The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (p -99.2) and the sample mean of any sample of size 5o is The z-score that locates the mean number of seats in the Delaware tourism board's sample (M-103.4) in the distribution of sample means is Use the unit normal tables and accompanying figures to answer the question that follows. To use the tables, click on the Unit Normal Tables tab below the figures, and use the dropdown box to select the desired range of z-score values. A table of the proportions of the normal distribution corresponding to that range of z-scores will appear. If you need a different range of z-scores, simply click on the box again and select a new range. Suggestion: Make a sketch of the area under the normal distribution you are seeking. This sketch will help you determine which column(s) of the norrmal table to use in determining the appropriate probability D:Between and z Body B Tail C The Unit Normal Tabes The Delaware tourism board selected a simple random sample of 50 full-service restaurants located within the state. Considering all possible such samples with n- 50, what is the probability of selecting one whose mean is greaber than 103.47 That is, p(M> 103.4) Gre

Answer 1

SufRere, X, X2, X50 a ramclem amfle df bouladien hawing m 50em a. Here, Xi i=1()50) densdes IRe ne neats in a rentaunant. So, tke nanden ample cemeso jL and vaniamce p2. bapulation wsilh meam E i nomfle aiee) Naturatly E (*) EE(Xi) E(x)(as Xin are - idendien namdem bariable and imdpendendhy distribucteol (99.2) &bb and v () - \asniance of namfle meam dintnibutien Xi v (x m LDue io imelufemdanse ) E V(xi) I Due do iderdlcal namdern uoriahler' Q

v(x.) 20.5 o niandancl of ihe dininiltion eJuen 20,5 5 0 50 Anaumingnoumal disisuibution op he Ramdom ariables X;, ue and nelve ike baeblem. may basceed TP Xi 99.2 m that UJe, atamdard eren, mRons uill be nome , i.e dinintbulion of Q.899i Vv x is a Ramdgm Here, we nee hat uarlable uth mean gq.Q amd ntamdrurdl dleuiatlon .P99 e neatr in e nomfle of 50 nentowrarits Tha io, en^ected qq.A amd obneruecl mRan m dyRical auerage dferane eseem iAe meam me. aomfie o ene n

Ae nample in (104- 99.) 4.2 namclomly nelucied nentaurani im dhoi f 99. ) han meam byhisal auenaye Similuuy deperence elueem he mean mumber o the ntuncland on neusice nestauram full neads i he 55 imDeiaiware and aamle 0. (ama Wemeu dininikulien of aamfie meam han mean 94.2 amd aiavdarudl erren we femous ualue of ullneRice nesdnnonds ditbution of aomle meam in 448(am