7: Sampling Distributions

Transcription

1 7: Samplig Distributios 7.1 You ca select a simple radom sample of size = 2 usig Table 1 i Appedix I. First choose a startig poit ad cosider the first three digits i each umber. Sice the experimetal uits have already bee umbered from to 999, the first 2 ca be used. The three digits OR the (three digits 5) will idetify the proper experimetal uit. For example, if the three digits are 742, you should select the experimetal uit umbered = 242. The probability that ay three digit umber is selected is 2 1 = 1 5. Oe possible selectio for the sample size = 2 is Each studet will obtai a differet sample, usig Table 1 i Appedix I. 7.3 Each studet will obtai a differet sample, usig Table 1 i Appedix I. 7.4 The studet should use the same procedure as i Exercise 7.1 ad refer to Table 1, Appedix I. 7.5 If all of the tow citizery is likely to pass this corer, a sample obtaied by selectig every teth perso is probably a fairly radom sample. 7.6 The questioaires that were retured do ot costitute a represetative sample from the 1 questioaires that were radomly set out. It may be that the voters who chose to retur the questioaire were particularly adamat about the Parks ad Recreatio surcharge, while the others had o strog feeligs oe way or the other. The orespose of half the voters i the sample will udoubtedly bias the resultig statistics. 7.7 Voter registratio lists ad DMV records will systematically exclude a large segmet of the geeral populatio odrivers, people who do ot ow cars, ovoters ad so o. 7.8 The wordig of the questio is biased to suggest that a yes respose is the correct oe. A more ubiased way to phrase the questio might be: Is there too much sex ad violece durig prime TV viewig hours? 7.9 Use a radomizatio scheme similar to that used i Exercise 7.1. Number each of the 5 rats from 1 to 5. To choose the 25 rats who will receive the dose of MX, select 25 two-digit radom umbers from Table 1. Each two-digit umber OR the (two digits 5) will idetify the proper experimetal uit. 7.1 a Sice the questio is particularly sesitive to people of differet ethic origis, you may fid that the aswers may ot always be truthful, depedig o the ethicity of the iterviewer ad the perso beig iterviewed. b Notice that the percetage i favor of affirmative actio icreases as the ethic origi of the iterviewer chages from Caucasia to Asia to Africa-America. The people beig iterviewed may be chagig their respose to match what they perceive to be the respose which the iterviewer wats to hear a The sample was chose from patiets i the cliical practice of oe of the authors who were willig to participate. This sample is ot radomly selected; it is a coveiece sample. b Valid ifereces ca be made from this study oly if the coveiece sample chose by the researcher behaves like a radom sample. That is, the patiets i this particular cliical practice must be represetative of the populatio of patiets as a whole. 164

2 c I order to icrease the chaces of obtaiig a sample that is represetative of the populatio of patiets as a whole, the researcher might try to obtai a larger base of patiets to choose from. Perhaps there is a computerized database from which he or she might select a radom sample a Sice each patiet must be radomly assiged to either aspiri or the experimetal drug with equal probability, assig the digits -4 to the aspiri treatmet, ad the digits 5-9 to the experimetal drug treatmet. As each patiet eters the study, choose a radom digit usig Table 1 ad assig the appropriate treatmet. b The radomizatio scheme i part a does ot guaratee a equal umber of patiets i each group a The first questio is more ubiased. b Notice that the percetage favorig the ew space program drops dramatically whe the phrase spedig billios of dollars is added to the questio Aswers will vary. May of the questios used i this policy survey were biased towards the political views held by members of the Republica Party Follow the istructios i the My Persoal Traier sectio. The blaks have bee filled i below. The samplig distributio of x will be approximately ormal with mea 53 ad stadard deviatio (or stadard error) Follow the istructios i the My Persoal Traier sectio. The blaks have bee filled i below. To fid the probability that the sample mea is greater tha 55, write dow the evet of iterest Px> ( 55). Whe x = 55, z = = =.67 σ / 3 Fid the probability: Px ( > 55) = Pz ( >.67) = = Follow the istructios i the My Persoal Traier sectio. The blaks have bee filled i below. The samplig distributio of x will be approximately ormal with mea 1 ad stadard deviatio (or stadard error) Follow the istructios i the My Persoal Traier sectio. The blaks have bee filled i below. To fid the probability that the sample mea is betwee 15 ad 11, write dow the evet of iterest P(15 < x < 11). Whe x = 15 ad x = 11, z = = = 1.58 ad z = = = 3.16 σ / 3.16 σ / 3.16 Fid the probability: P(15 < x < 11) = P(1.58 < z < 3.16) = = Regardless of the shape of the populatio from which we are samplig, the samplig distributio of the sample mea will have a mea µ equal to the mea of the populatio from which we are samplig, ad a stadard deviatio equal to σ. a µ = 1; σ = 3 36 =.5 b µ = 5; σ = 2 1 =.2 c µ = 12; σ = 1 8 = a If the sampled populatios are ormal, the distributio of x is also ormal for all values of. 165

3 b The Cetral Limit Theorem states that for sample sizes as small as = 25, the samplig distributio of x will be approximately ormal. Hece, we ca be relatively certai that the samplig distributio of x for parts a ad b will be approximately ormal. However, the sample size is part c, = 8, is too small to assume that the distributio of x is approximately ormal a The sketch of the ormal distributio with mea µ = 5 ad stadard deviatio σ =.2 is left to the studet. The iterval 5 ±.4or 4.6 to 5.4 should be located o the x axis. b The probability of iterest P(.15 < ( x µ ) <.15)ad is show below. c (.15 ( µ ).15) ( x µ ) P < x < = P < <.2 σ.2 = P.75 < <.75 = =.5468 ( z ) For a populatio with σ = 1, the stadard error of the mea is σ = 1 The values of σ for various values of are tabulated below ad plotted below. Notice that the stadard error decreases as the sample size icreases SE( x) = σ SE

4 7.24 a If the sample populatio is ormal, the samplig distributio of x will also be ormal (regardless of the sample size) with mea 16 µ = ad stadard deviatio (or stadard error) give as b Calculate σ = = z = = = 1.67, so that σ 2.4 P( x ) P( z ) c P( x ) P( z ) > 11 = > 1.67 = = < < 11 = 1.67 < < 1.67 = = a Age of equipmet, techicia error, techicia fatigue, equipmet failure, differece i chemical purity, cotamiatio from outside sources, ad so o. b The variability i the average measuremet is measured by the stadard error, σ. I order to decrease this variability you should icrease the sample size The weight of a package of 12 tomatoes is the sum of the 12 idividual tomato weights. Hece, sice weights ad heights are i geeral ormally distributed, so will be the sum of the 12 weights, accordig to the Cetral Limit Theorem The umber of bacteria i oe cubic foot of water ca be thought of as the sum of 1728 radom variables, each of which is the umber of bacteria i a particular cubic ich of water. Hece, the Cetral Limit Theorem isures the approximate ormality of the sum a Sice the sample size is large, the samplig distributio of x will be approximately ormal with mea µ = 65, 68 ad stadard deviatio σ = 4 6 = b From the Empirical Rule (ad the geeral properties of the ormal distributio), approximately 95% of the measuremets will lie withi 2 stadard deviatios of the mea: ( ) µ ± 2SE 65, 68 ± , 68 ± or to 66, 64.8 c Use the mea ad stadard deviatio for the distributio of x give i part a. 67, 65, 68 P( x > 67,) = P z > = P z > 2.7 = =.35 ( ) d Refer to part c. You have observed a very ulikely occurrece, assumig that µ = 65,68. Perhaps your sample was ot a radom sample, or perhaps the average salary of $65,68 is o loger correct a The populatio from which we are radomly samplig = 35 measuremets is ot ecessarily ormally distributed. However, the samplig distributio of x does have a approximate ormal distributio, with mea µ ad stadard deviatio σ Sice. The probability of iterest is ( µ 1) ( 1 ( µ ) ) P x < = P < x <1. x µ z = has a stadard ormal distributio, we eed oly fid σ to approximate the above σ probability. Though σ is ukow, it ca be approximated by s = 12 ad σ = The ( µ < 1) = ( < < ) P( z ) P x P z =.49 < <.49 = =

5 b No. There are may possible values for x, the actual percet tax savigs, as give by the probability distributio for x. 7.3 a Sice the origial populatio is ormally distributed, the sample mea x is also ormally distributed (for ay sample size) with mea µ ad stadard deviatio σ 2 1 b If µ = 21ad = 1, the z-value correspodig to x = 2 is ad 2 21 z = = = 1.58 σ 2 1 ( ) P( z ) P x < 2 = < 1.58 =.571 c Refer to part b ad assume that µ is ukow. It is ecessary to fid a value µ so that ( 2 ).1 P x < =. Recall that if µ = µ, the z-value correspodig to x = 2 is µ 2 µ 2 µ x z = = =. σ µ 2 µ < = < = = It is ecessary the to have P( x ) P z A Refer to the figure below. From Table 3, the value of z that cuts off.1 i the left had tail of the ormal distributio is z = 3.8. Hece, the two values, ( µ ) ad z = 3.8, must be the same. Solvig for µ, 2 µ.632 = 3.8 or µ = a The radom variable T = xi, were xi is ormally distributed with mea µ = 63 ad stadard deviatio σ = 4 for i = 1, 2, 3. The Cetral Limit Theorem states that T is ormally distributed with mea µ = 3(63) = 189 ad stadard deviatio σ = 4 3 = b Calculate T z = = =

6 ( ) ( ) The PT> 2 = Pz< 1.59 = = a Sice the total daily sales is the sum of the sales made by a fixed umber of customers o a give day, it is a sum of radom variables, which, accordig to the Cetral Limit Theorem, will have a approximate ormal distributio. b Let x i be the total sales for a sigle customer, with i = 1, 2,., 3. The x i has a probability distributio with µ = 8.5 ad σ = 2.5. The total daily sales ca ow be writte as x = xi. If = 3, the mea ad stadard deviatio of the samplig distributio of x are give as µ = 3(8.5) = 255 adσ = = a Sice the origial populatio is ormally distributed, the sample mea x is also ormally distributed (for ay sample size) with mea µ ad stadard deviatio σ.8 13 =.716 The z-value correspodig to x = is z = = = 4.99 σ.8 13 ad P x < = P z < 4.99 ( ) ( ) b Sice the probability is extremely small, the average temperature of degrees is very ulikely The sampled populatio has a mea of 5.97 with a stadard deviatio of a With = 31, calculate z = = = 1.51, so that σ 1.95/ 31 ( ) P( z ) P x 6.5 = 1.51 =.9345 b Calculate z = = = 1.94, so that σ 1.95/ 31 ( ) P( z ) P x 9.8 = = c The probability of observig a average diameter of 9.8 or higher is extremely ulikely, if ideed the average diameter i the populatio of affected tedos was o differet from that of uaffected tedos (5.97). We would coclude that the average diameter i the populatio of patiets affected with AT is higher tha Follow the istructios i the My Persoal Traier sectio. The blaks have bee filled i below. The samplig distributio of stadard error) pq.7(.3) = 5 =.648. ˆp will be approximately ormal with mea p =.7 ad stadard deviatio (or 7.36 Follow the istructios i the My Persoal Traier sectio. The blaks have bee filled i below. To fid the probability that the sample proportio is less tha.8, write dow the evet of iterest P( p ˆ <.8). Whe p ˆ =.8, pˆ p.8.7 z = = = 1.54 pq.648 Fid the probability: P( pˆ <.8) = P( z < 1.54) =

7 7.37 a p SE( pˆ ) pq.3(.7) =.3; = = = pq.1(.9) =.1; = = =.15 4 b p SE( pˆ ) pq.6(.4) =.6; = = = c p SE( pˆ ) 7.38 Sice the sample sizes are very large, the samplig distributios i Exercise 7.37 will each be approximately ormal, with appropriate meas ad variaces. The iterval p± 2SE pˆ should cover 95% of the measuremets a-b The probability of iterest is P pˆ p.8 = P.8 pˆ p.8 Sice ( ) ( ( ) ) ˆp is approximately ormal, with stadard deviatio ( ).8.8 P(.8 ( pˆ p).8) = P z = P = =.9198 ( z ) ( ) SE p ˆ =.458 from Exercise 7.37, 7.4 a For = 5 ad p =.1, p = 5 ad q = 45 are both greater tha 5. Therefore, the ormal approximatio will be appropriate. b ( ).12 ˆ.12.1 P p > = P z > = P( z > 1.49) = = c ( ).1 ˆ.1.1 P p < = P z < = P( z < ) = d P( ( pˆ p) ) P( z ).2.2 = = = The values SE = pq for = 1 ad various values of p are tabulated ad graphed below. Notice that SE is maximum for p =.5 ad becomes very small for p ear zero ad oe. p SE( p ˆ) SE p

8 7.42 a For = 4 ad p =.8, p = 32 ad q = 8 are both greater tha 5. Therefore, the ormal pq.8(.2) approximatio will be appropriate with SE = = =.2. 4 b ( ).83 ˆ.83.8 P p > = P z > = P( z > 1.5) = = c P(.76 < pˆ <.84) = P < z < = P( 2 < z < 2 ) = = a For = 1 ad p =.46, p = 46 ad q = 54 are both greater tha 5. Therefore, the ormal pq.46(.54) approximatio will be appropriate, with mea p =.46 ad SE = = = b ( ).5 ˆ.5.46 P p > = P z > = P( z >.8) = = c P(.35 < pˆ <.55) = P < z < = P( 2.21 < z < 1.81 ) = = d The value p ˆ =.3 lies pˆ p.3.46 z = = = 3.21 pq.498 stadard deviatios from the mea. This is a ulikely occurrece, assumig that p =.46. Perhaps the samplig was ot radom, or the 46% figure is ot correct a The radom variable ˆp, the sample proportio of studets who used the Iteret as a major resource i the past year, has a biomial distributio with = 1 ad p =.66. Sice p = 66 ad q = 34 are both greater tha 5, this biomial distributio ca be approximated by a ormal distributio with mea.66(.34) p =.66 ad SE = = b ( ).68 ˆ P p > = P z > = P( z > 1.33) = = P.64 < pˆ <.68 = P 1.33 < z < 1.33 = =.8164 c ( ) ( ) d The value p ˆ =.7 lies pˆ p.7.66 z = = = 2.67 pq.15 stadard deviatios from the mea. This is a somewhat ulikely occurrece, assumig that p =.66, ad would ted to cotradict the reported figure a The radom variable ˆp, the sample proportio of brow M&Ms i a package of = 55, has a biomial distributio with = 55 ad p =.13. Sice p = 7.15 ad q = are both greater tha 5, this biomial distributio ca be approximated by a ormal distributio with mea p =.13 ad.13(.87) SE = = b ( ).2 ˆ.2.13 P p < = P z < = P( z < 1.54 ) =

9 ˆ > = > = > =.4535 d From the Empirical Rule (ad the geeral properties of the ormal distributio), approximately 95% of the measuremets will lie withi 2 (or 1.96) stadard deviatios of the mea: p± 2 SE.13 ± 2(.4535) c P( p ) P z P( z ).13 ±.9 or.4 to The radom variable ˆp, the sample proportio of overweight childre i a radom sample of = 1, has a biomial distributio with = 1 ad p =.15. Sice p = 15 ad q = 85 are both greater tha 5, this biomial distributio ca be approximated by a ormal distributio with mea p =.15 ad.15(.85) SE = = a ( ).25 ˆ P p > = P z > = P( z > 2.8) = = b ( ).12 ˆ P p < = P z < = P( z <.84) = c ( ).3 ˆ.3.15 P p > = P z > = P( z > 4.2) 1 1 =. Sice the z-score is so high ad the probability.3571 is so low, it would be highly ulikely to fid 3% of the childre to be overweight a The radom variable ˆp, the sample proportio of cosumers who like uts or caramel i their chocolate, has a biomial distributio with = 2 ad p =.75. Sice p = 15 ad q = 5 are both greater tha 5, this biomial distributio ca be approximated by a ormal distributio with.75(.25) mea p =.75 ad SE = = b ( ).8 ˆ.8.75 P p > = P z > = P( z > 1.63) = = c From the Empirical Rule (ad the geeral properties of the ormal distributio), approximately 95% of the measuremets will lie withi 2 (or 1.96) stadard deviatios of the mea: p± 2 SE.75 ± 2(.362).75 ±.6 or.69 to a The upper ad lower cotrol limits are s.87 UCL = x + 3 = = = s.87 LCL = x 3 = = = b Cotrol charts are used to moitor the process variable, detectig shifts that might idicate cotrol problems. c The cotrol chart is costructed by plottig two horizotal lies, oe the upper cotrol limit ad oe the lower cotrol limit (see Figure 7.15 i the text). Values of x are plotted, ad should remai withi the cotrol limits. If ot, the process should be checked Similar to Exercise a The upper ad lower cotrol limits are 172

10 s 4.3 UCL = x + 3 = = = s 4.3 LCL = x 3 = = = b The cotrol chart is costructed by plottig two horizotal lies, oe the upper cotrol limit ad oe the lower cotrol limit (see Figure 7.15 i the text). Values of x are plotted, ad should remai withi the cotrol limits. If ot, the process should be checked. 7.5 The x chart is used to moitor the average value of a sample of quatitative data, while the p chart is used to moitor qualitative data by coutig the umber of defective items ad trackig the percetage defective a The upper ad lower cotrol limits for a p chart are p(1 p).35(.965) UCL = p + 3 = = =.9 1 p(1 p).35(.965) LCL = p 3 =.35 3 = =.2 1 or LCL = (sice p caot be egative). b The cotrol chart is costructed by plottig two horizotal lies, oe the upper cotrol limit ad oe the lower cotrol limit (see Figure 7.16 i the text). Values of ˆp are plotted, ad should remai withi the cotrol limits. If ot, the process should be checked The upper ad lower cotrol limits for a p chart are p(1 p).41(.959) UCL = p + 3 = = =.83 2 p(1 p).41(.959) LCL = p 3 =.41 3 = =.1 2 or LCL = (sice p caot be egative). b The cotrol chart is costructed by plottig two horizotal lies, oe the upper cotrol limit ad oe the lower cotrol limit (see Figure 7.16 i the text). Values of ˆp are plotted, ad should remai withi the cotrol limits. If ot, the process should be checked a The upper ad lower cotrol limits are s 165 UCL = x + 3 = 1, = 1, = 12, s 165 LCL = x 3 = 1, = 1, = b The x chart will allow the maager to moitor daily gais or losses to see whether there is a problem with ay particular table The upper ad lower cotrol limits for a p chart are p(1 p).21(.979) UCL = p + 3 = = =.43 4 p(1 p).21(.979) LCL = p 3 =.21 3 = =.1 4 or LCL = (sice p caot be egative). The maager ca use the cotrol chart to detect chages i the productio process which might produce a uusually large umber of defectives. 173

11 pˆ Calculate i + + L+ p = = =.197. The upper ad lower cotrol limits for the p chart are the k 3 p(1 p).197(.83) UCL = p + 3 = = = p(1 p).197(.83) LCL = p 3 = = = The upper ad lower cotrol limits are UCL = x + 3 s.7 = = = LCL = x 3 s.7 = = = Usig all 14 measuremets, the value of s is calculated to be s = ad x =.256. The the upper ad lower cotrol limits are UCL = x + 3 s = = LCL = x 3 s = = The upper ad lower cotrol limits are s.264 UCL = x + 3 = = = s.264 LCL = x 3 = = = The cotrol chart is costructed by plottig two horizotal lies, oe the upper cotrol limit ad oe the lower cotrol limit (see Figure 7.15 i the text). Values of x are plotted, ad should remai withi the cotrol limits. If ot, the process should be checked Refer to Exercise The sample mea is outside the cotrol limits i hours 2, 3 ad 4. The process should be checked. 4 4! 7.6 a C 2 = = 6 samples are possible. 2!2! b-c The 6 samples alog with the sample meas for each are show below. Sample Observatios x 1 6, , , , , , d Sice each of the 6 distict values of x are equally likely (due to radom samplig), the samplig distributio of x is give as 1 p( x) = for x = 1.5,2, 2.5,3.5, 4, The samplig distributio is show o the ext page. 174

12 p(x) x e The populatio mea is ( ) 4 3 a value of x exactly equal to the populatio mea. µ = =. Notice that oe of the samples of size = 2 produce 7.61 Refer to Exercise 7.6. If samples of size = 3 are draw without replacemet, there are 4 possible samples with sample meas show below. Sample Observatios x 1 6, 1, , 1, , 3, , 3, 2 2 The samplig distributio of x is the 1 p( x) = 4 for x = 2,3,3.333,3.667 The samplig distributio is show below. p(x).25 2 x a The distributio of lead cotet readigs for idividual water specimes is probably skewed to the right, with a few specimes cotaiig a very large amout of lead. This coclusio is cofirmed by lookig at the mea ad stadard deviatio,.33 ad.1, respectively. The large stadard deviatio does ot allow the measuremets to spread over the rage µ ± 2σ without ragig ito egative levels (which are impossible). 175

13 b Sice the sample of 23 daily lead levels is draw from a populatio of meas based o = 4 observatios, the sample is beig draw from a approximately ormal populatio, accordig to the Cetral Limit Theorem. c For the sample meas i part b, the samplig distributio has mea µ =.33 ad stadard deviatio σ =.1 4 = a Usig the rage approximatio, the stadard deviatio σ ca be approximated as R 55 5 σ = = b The samplig distributio of x is approximately ormal with mea µ ad stadard error σ =.625 The ( 2) 2 2 P = P z = P( 3.2 z 3.2 ) = = c If the scietists are worried that the estimate of µ = 35 is too high, ad if your estimate is x = 31.75, the your estimate lies z = = = 5.2 σ.625 stadard deviatios below the mea. This is a very ulikely evet if i fact µ = 35. It is more likely that the scietists are correct i assumig that the mea is a overestimate of the mea biomass for tropical woodlads It is give that = 3, µ = 9 ad σ = 1. The P [ sample mea satisfies ANSI] = P( x < 85) 85 9 = P < = P < σ 1 3 =.1922 ( z.87) 7.65 a To divide a group of 2 people ito two groups of 1, use Table 1 i Appedix I. Assig a idetificatio umber from 1 to 2 to each perso. The select te two digit umbers from the radom umber table to idetify the te people i the first group. (If the umber is greater tha 2, subtract multiples of 2 from the radom umber util you obtai a umber betwee 1 ad 2.) b Although it is ot possible to select a actual radom sample from this hypothetical populatio, the researcher must obtai a sample that behaves like a radom sample. A large database of some sort should be used to esure a fairly represetative sample. c The researcher has actually selected a coveiece sample; however, it will probably behave like a simple radom sample, sice a perso s ethusiasm for a paid job should ot affect his respose to this psychological experimet a Sice the data already existed before the researcher decided to study it, this is a observatioal study. b Sice the subject of the study is a sesitive oe, there will be problems of orespose ad/or iaccurate resposes to the questios a From the 5 lettuce seeds, the researcher must choose a group of 26 ad a group of 13 for the experimet. Idetify each seed with a umber from 1 to 5 ad the select radom umbers from Table 1. The first 26 umbers chose will idetify the seeds i the first Petri dish, ad the ext 13 will idetify the seeds i the third Petri dish. If the same umber is picked twice, simply igore it ad go o to the ext umber. Use a similar procedure to choose the groups of 26 ad 13 radish seeds. b The seeds i these two packages must be represetative of all seeds i the geeral populatio of lettuce ad radish seeds a This is a observatioal study, sice the data existed before you decided to observe or describe it. 176

14 b The researcher should be cocered about orespose ad utruthful resposes, due to the sesitive ature of the questio Referrig to Table 1 i Appedix I, we will select 2 umbers. First choose a startig poit ad cosider the first four digits i each umber. If the four digits are a umber greater tha 7, discard it. Cotiue util 2 umbers have bee chose. The customers have already bee umbered from 1 to 7. Oe possible selectio for the sample size = 2 is a For this biomial radom variable with = 5 ad p =.85, the mea ad stadard deviatio of ˆp are µ = p =.85 ad SE =.85(.15) = b Sice p = 5(.85) = 425 ad q = 75 are both greater tha 5, the ormal approximatio is appropriate. c ( ) ˆ P p > = P z > = P( z > 1.88) = 1.31 = d P(.83 < pˆ <.88) = P < z < = P( 1.25 < z < 1.88 ) = = e For a ormal (or approximately ormal) radom variable, the iterval µ ± 2.58σ will cotai 99% of the measuremets. For this biomial radom variable x, this iterval is p± 2.58 SE.85 ± 2.58(.1597).85 ±.4 or.81 to a Sice each cluster (a city block) is cesused, this is a example of cluster samplig. b This is a 1-i-1 systematic sample. c The wards are the strata, ad the sample is a stratified sample. d This is a 1-i-1 systematic sample. e This is a simple radom sample from the populatio of all tax returs filed i the city of Sa Berardio, Califoria Defie x i to be the weight of a particular ma or woma usig the elevator. It is give that x i is approximately ormally distributed with µ = 15 ad σ = 35. The accordig to the Cetral Limit Theorem, the sum or total weight,, will be ormally distributed with mea µ = 15 ad stadard deviatio σ = 35 The z-value correspodig to so that we eed xi. It is ecessary to fid a value of such that P( x i > 2 ) =.1 x i = 2 is xi µ 2 15 z = = σ P z > =.1 35 From Table 3, we coclude that 2 15 = Maipulatig the above equatio, we obtai a 35 quadratic equatio i. 177

15 ( ) 2 15 = ,, 66, , 5 = Usig the quadratic formula, the ecessary value of is foud. 2 b± b 4ac 66, ± 89, = = 2a 45, Choosig the smaller root of the equatio, 517, = = , Hece, = 12 will provide a sample size with P( x i > 2 ).1. I fact, for = 11, while for = 12, ( ) ( ) P x > 2 = P z > 3.2 = =.13 i ( ) ( ) P x > 2 = P z > 1.65 = =.495 i 7.73 a The umber of packages which ca be assembled i 8 hours is the sum of 8 observatios o the radom variable described here. Hece, its mea is µ = 8(16.4) = ad its stadard deviatio is σ = = b If the origial populatio is approximately ormal, the samplig distributio of a sum of 8 ormal radom variables will also be approximately ormal. Sice the origial populatio is exactly ormal, so will be the samplig distributio of the sum. c ( ) P x > = P z > = P( z > 1.3) = = a The total productio is the sum of 1 observatios o the radom variable x described i Exercise Hece, its mea is 1( 131.2) = 1312 ad its stadard deviatio is = productio < = < = < 2.75 = b P[ ] P z P( z ) 7.75 a The average proportio of defectives is L+.3 p = = ad the cotrol limits are p(1 p).32(.968) UCL = p + 3 = = p(1 p).32(.968) ad LCL = p 3 =.32 3 =.28 1 If subsequet samples do ot stay withi the limits, UCL =.848 ad LCL =, the process should be checked. b From part a, we must have p ˆ >.848. c A erroeous coclusio will have occurred if i fact p <.848 ad the sample has produced p ˆ =.15 by chace. Oe ca obtai a upper boud o the probability of this particular type of error by calculatig P pˆ.15 whe p =.848. ( ) 178

16 7.76 The probability that a shipmet of 1 bulbs will have o more tha 4% defective is P( pˆ.4). Sice ˆp p(1 p).32(.968) is approximately ormal with mea p.32 ad SE = =.176, P( pˆ.4) = P z = P( z.45) = Refer to Exercise 7.75, i which UCL =.848 ad LCL =. For the ext 5 samples, the values of ˆp are.2,.4,.9,.7,.11. Hece, samples 3 ad 5 are producig excess defectives. The process should be checked L a If the process is i cotrol, σ = 1.2. Calculate x = = = With = 5, the upper ad lower cotrol limits are s 1.2 x ± 3 = ± 3 = ± or LCL = ad UCL = b The ceterlie is x = ad the graph is omitted. Oly oe sample exceeds the UCL; the process is probably i cotrol Aswers will vary from studet to studet. Payig cash for opiios will ot ecessarily produce a radom sample of opiios of all Pepsi ad Coke drikers. 7.8 Aswers will vary. Oe solutio is to umber the cotaiers from 1 to 2. Refer to Table 1 i Appedix I, ad select 2 two-digit umbers, choosig a radom startig poit. If the two digit umber is greater tha 2, subtract 2 from the umber cosecutively, util you have a umber betwee 1 ad 2. Cotiue util 1 umbers have bee chose. These te cotaiers will be stored at oe temperature; the other te will be stored at the secod temperature a Sice the fill per ca is ormal with mea 12 ad stadard deviatio.2, the total fill for a pack of six cas will also have a ormal distributio with mea 24(12) 288 µ = = ad stadard deviatio σ =.2 24 =.9798 b Let T be the total fill for the case of soda. The PT ( < 286) = Pz ( < ) = Pz ( < 2.4) = c Px ( < 11.8) = Pz ( < ) = Pz ( < 2.45) =.71.2 / Similar to Exercise Sice the package weights are ormal with mea 16 ad stadard deviatio.6, the total weight for a box of 24 packages will also have a ormal distributio with mea µ = 24(16) = 384 ad stadard deviatio σ =.6 24 = Let T be the total weight for the box. The PT ( > 392) = Pz ( > ) = Pz ( > 2.72) = = a The average proportio of ioperable compoets is L p = = =.1 5(15) 75 ad the cotrol limits are 179

17 p(1 p).1(.9) UCL = p + 3 = = p(1 p).1(.9) ad LCL = p 3 =.1 3 = If subsequet samples do ot stay withi the limits, UCL =.2273 ad LCL =, the process should be checked a The radom variable x has a discrete probability distributio give as p( x) Usig the formulas give i Chapter 4, 1 21 µ = xp( x) = ( L+ 6) = = σ = x p( x) µ = = = ( L ) ( ) 2 1 = for x = 1,2,3,4,5,6. 6 ad σ = = 1.78 b-c Aswers will vary from studet to studet. The distributio should be relatively uiform with mea ad stadard deviatio close to those give i part a a The theoretical mea ad stadard deviatio of the samplig distributio of x whe = 2 are µ = 3.5 ad SE = σ = = 1.28 b-c Aswers will vary from studet to studet. The distributio should be relatively uiform with mea ad stadard deviatio close to those give i part a a The theoretical mea ad stadard deviatio of the samplig distributio of x whe = 3 are µ = 3.5 ad SE = σ = =.986 b-c Aswers will vary from studet to studet. The distributio should be relatively uiform with mea ad stadard deviatio close to those give i part a a The theoretical mea ad stadard deviatio of the samplig distributio of x whe = 4 are µ = 3.5 ad σ = =.854 b-c Aswers will vary from studet to studet. The distributio should be relatively uiform with mea ad stadard deviatio close to those give i part a a If the sample populatio is ormal, the samplig distributio of x will also be ormal (regardless of the sample size) with mea 1 µ = ad stadard deviatio (or stadard error) give as SE = σ =.36 5 =.161 b Use the Normal Probabilities for Meas applet. Eter the values of µ, σ, (the warig remids you that x must be ormal) ad x = 1.3, choosig Oe-tailed from the dropdow list. The applet shows P x > 1.3 =.312. If you choose to use had calculatios, calculate ( ) ( ) P( z ) z = = = 1.86, so that σ.161 P x > 1.3 = > 1.86 = =.314 (the applet uses full decimal accuracy) 18

18 c Chage the applet to x =.5 ad Cumulative from the dropdow list. The applet shows 4 ( ) ( ) P x <.5 = 9. 1 =.9. If you choose to use had calculatios, calculate.5 1 z = = = 3.11, so that σ.161 ( ) P( z ) P x <.5 = < 3.11 =.9 d Chage the applet to x = 1.4 ad Two-tailed from the dropdow list. The probability that x deviates from µ = 1by more tha.4 is the probability that x exceeds 1.4 or is less tha.6. The applet shows ( ) P( x ) P x > <.6 =.13. If you choose to use had calculatios, calculate x = = = 2.48 ad z σ µ z x µ = = = The P( x ) P( x ) P( z ) P( z ) ( ) σ 2.48 > <.6 = > < 2.48 = 2.66 = Use the Normal Probabilities for Meas applet. Eter the values of µ, σ,. a Eter 11 x =, choosig Cumulative from the dropdow list. The applet shows ( ) P x < 11 =.62. Sice the etire area to the left of 111 is.5, the area betwee 11 ad 111 is.5.62 = b Eter 112 P x > 112 =.62. c Eter 9 x =, choosig Oe-Tailed from the dropdow list. The applet shows ( ) x =, choosig Cumulative from the dropdow list. The applet shows P( x < ) = 9.. Case Study: Samplig the Roulette at Mote Carlo 1 Each bet results i a gai of ( $5) if he loses ad $175 if he wis. Thus, the probability distributio of the gai x o a sigle $5 bet is x p(x) 5 37/ /38 2 The ( ) xp x ( )( ) ( ) E x = ( ) = = x ( ) ( ) ( ) ( ) ( ) σ = x p( x) µ = = The gai for the eveig is the sum S = xi of the gais or losses for the 2 bets of $5 each. Whe the sample size is large, the Cetral Limit Theorem assures that this sum will be approximately ormal with mea µ = µ = = $52.64 ad variace S 2 2 S x ( ) σ = σ = 2( ) = 166, σ = 166, = S The total wiigs will vary from $1 (if the gambler loses all 2 bets) to $35, (if the gambler wis all 2 bets), a rage of $36,. However, most of the wiigs (95%) will fall i the iterval µ ± 2σ = ± 2(47.43) or $867.5 to $ The large gais are highly improbable. S S 181

19 4 The loss of $1 o ay oe ight will occur oly if there are o wis i 2 bets of $5. The probability 2 37 of this evet is =.5. Defie y to be the umber of eveigs o which a loss of $1 occurs. The y has 38 a biomial distributio with p =.5 ad = 365. Usig the Poisso approximatio to the biomial with µ = p = 1.825, the probability of iterest is approximately which is highly improbable. ( ) e p(7) =.2 7! 5 The largest eveig s wiigs, $116, is ot surprisig. It lies ( ) z = = stadard deviatios above the mea, so that P wiigs 116 = P z 2.98 = =.14 ( ) ( ) for ay oe eveig. The probability of observig wiigs of $116 or greater o oe eveig out of 365 is the approximated usig the Poisso approximatio with 365(.14).511 µ = = or ( ) e p(1) =.365 1! 182