Chap 2 answers

The Practice of Statistics: TI-83/89 Graphing Calculator Enhanced

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Unformatted text preview: The Normal Distributions 2.1 There are many correct drawings. Here are two possibilities: {a} {b} 2.2 {a} The area under the eurve is a rectangle with height 1 and width 1. Thus the total area under the curve =1 ><l=1_ [b] 23%. [The region is a rectangle with height 1 and base 1Width D2; hence the area is {1.2.} {e} 6U 02b. [d] 511%. {e} Mean = 1X; or [15, the “balanee point” of tile densityr curve. 2.3 {a} Total area under curve = area of triangle + area of 2 rectangles = lfzfilth] + {Ail} + {511(1) =fl.2 + fl.4+{i.4 = 1. age; nine {a} 0.33. {e} The median is the “equalrareas’” point. By {d}, the area between I} and :12 is [1.35. The are between 114 and [L3 is 0.4. Thus the “equalaareas” point must lie between 11.2 and [1.4. ' 2.4 {a} Mean C, median B; {b} mean A, median Pi; {Cl mean A. median B. 2.5 The uniform distribution. Eaeh of the 6 bars should have a height of 23. 2.45 2.3 Chapter 2 2.2 {a} 2.5% {this is 2 standard deviations above the mean}. {b} 69 i 5; that is, 64 to 24 inches. {c} 1696. [d] 3491:. The area to the left of X = 21.5 under the NE69, 2.5] curve is [1.84. as [a]5[19t+.{b)2.59t.[c}lll} : so, or so to 15s. 2.9 {a} 56. The mean [center of the distribution in this case) 64.5 under the N{64.5, 2.5) curve. Parts [13} through {d} use si 2.1[1 Ansyvers vary. is 64.5, so 512196 of the area to the left of milar reasoning. {b} 2.5 {c} 84 {d} 99.35. 2.11 Approximately [1.2 [for the tall one} and [1.5. 2.12 {a} 1696. {b} 34th percentile. The area to the left of X = 23.9 under the 19122.3, 1.1} curve is 0.34. {c} 613%. 2.13 [a1266 i 32, or 234 to 2913 days. I: deviations to the right of the mean}. 2.14 [a] By fl'le 63-95992 rule, approximately 1696 of the scores lie below ,u. - lo' = 111.1 — 25 =- 85. The T163 command shadeNorm {—1090, 35, 1119, 25} reports a lower left tail area of .153655. [bl The 34th percentile is the area under the N{11|L1,25} curve to the left of p. + lo- = MD + 25 = 135. The command shachorm {— 1111311, 135, 119, 25} reports an area of .341345. The 92.5 percentile is the area to the left oftt + 2::- = 11[1 + 50 = 1611. ShadeNorm {—1901}, 16[1, 1113,25} reports an area of .92225. 2-1 s {a} b} Less than 234 days. [c] More than 298 days {2 standard 4s? sss sss ss.1 sea :22: 22.5 ts—acrtt—Zcr‘u—lrr p. at+laps+2erp+5a [bile-39111583, 62.9}. 9596: {53.5, 2221992964482, 22.5}._ 2.16 [a] ball—- {b} 56% of the outcomes are 1 median is 1, Q] = 5? and Q3 2 [1.13 {122} = 11.4. 2.12 {a} [)utcomes'around 25 are more likely. [{1} with center at about 25, single peaked at the cent outliers. The normal density curve should fit this The distribution should be roughly symmetric er, standard deviation about 3.5, and few or no histogram well. 1111! Normal Distributions 29 2.13 The appearancc cf the graph indicates that the functions ‘1"[ and Y; are identical; they bnth represent the density curve of the standard ncrma] probability distribution. 2.19 Eleaner's z-sccre is [681} -- Silflfllfll} = 1.8; lGerald’s is [2? — 13}f6 = 1.5. Eleancr's score is higher. 2.21] Cobb: z = [.4213 — .2156} + .DSTI = 4.15; Williams: a = 4.26:; Brett: 2 = 4.0?. Williams’ 2— sccrc is highest. 2.21 {a} flaws. {b} 1 — casts; ancza D 2.35 0 2.35 {c} 1 — 13.11435 = 119515. {a} saws ~ cmss = 0.9493. —1.56 11 "1.65 1} 2.35 2.22 {a} About —c.sr5 {—c.s?44a}. [bi About 0.25 {assays}. ‘ A <61"? . .2533 5!} Chapter 2 2.23 {alz = 3f2.5 ——- 1.20; 1 - {1.3849 = Ill 151; about 11.5%. {b} About 33.5%. {c} 22.2 inches. 2.24 {a} 65.5%. {b} 5.5%. to} About 127'r [or more}. 2.25 {a} Area under the density curve of X = N{fl.3?. I164} to the right of {1.441 = .2266. {13] Area under the density curve of X = MUN. £1.64] between 6.4!] and [LED = .226]. {-2} Proportion in {a} changes to .6915; proportion in {h} changes to .6415. 2.2s {a} 48 8 45 5|] 'i‘ 51 El 52 6199 53 D4463 5-4 246'? '35 DEBTS 56 12353 5'? [19 58 5 {b} E = 5.4429 and s = 0.221195. About 25.8% [22 out of 29} lie within one standard deyiation of f, while 96.6% [28229] lie within two standard deviations. ll 11 The linearity of the normal probability plot indicates an approximately normal dis- tribution. The Normal Distrlhutiuns 3' 2.22 {a} 9 4 16 ll 12 12346 13 22225668 14 3629 15 232268 16 122446288 12 638 18 236?? 19 l? 20 21 22 8 The distribution is approximately symmetric with a peak at the 16 stern. Outliers exist at 9.4 and 22.8. {b} i = 15.536, Median = 15.25. The similarity of these measures reflects the symmetry observed in part (a)- "1, {e} i = 15.536, s = 2.55. About 63.2% {311 out of 44] fall within one standard deviation of E, and about 95.45% {42 out of 44} fall within two standard deviations of f . .411 data values [1|]l'l‘l-6, or 44 out of 44} fall within three standard deviations of f. 1 s 14 is r 1 {2.3%} {11.4%} (has; {35.4%} {15.9%} {2.36} 10.436 13.1136 15.586 13.156 211.636 E — 23 3E - s E i + s E + 2s {d} Window: [5. 2515 by [-3, 3h. The plot is strongly linear, with the smallest and largest length observations, 9.4 and 22.3, being the most significant deviations from linearity. [e] The data appear to be approximately normal. The data satisfy the 63-95-953 rule and die shape of the distribution is similar to that of a normal distribution. Also, the strong linearity of die normal probability plot suggests that the data were drawn from a normal distribution. 2.23 {a} 6.6122. [b] 121.9828. {4:} 13.6334. {a} 6.9494. 32 Chapter 2 2.29 {a} About 11.52. {b} About — 1.114. {c} About 11.24. {:1} About —1.2s. 2.30 {a} 12% i 2(16.5%} = -21% to 45%. {b} About 0.23: x s: 0% corresponds to z s: —0.23, for which Table A gives 0.2355 [software gives 0.2322}. {c} About 0.215: x 1: 25% corresponds to z :e 0.29, for which Table A gives 0.21411 {software gives 0.2154}. 2.31 {a} About 5.21%. {b} About 55%. {c} Approximately 220 days or longer. 2.32 {a} A score of x = 130 is z = 2 standard deviations above the mean, so about 2.5% of 1932 children had very superior scores {2.20% if using Table A}. {b} x :4 130 corresponds to .2 L? 0.62, for which Table A gives 0.2514—about 25%. 2.55 {a} At about i 0.625. {b} For dard deviations from the mean; 255.2 and 226.3. 2.34 {a} At about : 1.22. {b} 64.5 a 2.2, or 151.310 62.2. 2.35 {a} 1T: = 1.442, s = 0.3035, Median = 1.45. distribution is roughly symmetric. any normal distribution, the quartiles are i 0.625 stand for human pregnancies, the quartiles are 266 i 10.0, or The mean and median are virtually identical; the {b} Approximately 62% [20 out of 30} of all observations lie within one standard devia- tion of the mean. 100% of the observat ions lie within two and within three standard deviations. 0 5 10 10 5 0 {0%} {16.2%} {33.3%} {33.3%} {16.2%} {0%} 0.355 1.1385 1.442 1.2455 2.0451 f-2s f—s ':E §+s i+2s {c} The normal probability plot is strongly lineat, indicating that the data fit a normal dis- tribution well. 111a.I Normal Distributions 53 {d} The data appear to be approximately normal. The shape of the distribution is symmetric, the data approximately obey.r the fill-9539.? rule, and the normal probability plot suggests that a normal model is appropriate. 2.36 The normal probability plot shows a strong linear trend. The presidents' ages are 31313me matel}.f normally.r distributed. 2.3? The mean and standard deviation of a data set of standardized values should be 0 and 1 respectively. The command l-Var Stats LSTDSC produces the following screen: 1:Uar Stats x=3.3?29935'15 Ex=3.EE'13 Ex3=42 5x=1 ox=.9333836912 i ¢n=43 a and confirms that the mean is I] and the standard deviation is 1. 2.33 {a} 51!: Chapter 2 [b] If X is the coordinate for which the area between 4:] and X under the curve is .5, then 13“.? X base X height = FIE = U2. Solving for X: X = l. The median is l. The same approach shows that Q] =‘Vfi = .TD? and Q3 = V1.5 = 1.225. {c} The mean will lie to the left of the median [1} because the density curve is skewed left. {d} Area = [.5] {.EJEZ = .125, so 12.5% of the observations lie below [3.5. None [(1%) of the observations lie_‘ above 1.5. 2.39 {a} {b} M = as. Q. = .s, g. = .s. {a 25.2%. is} 49.6%. 2.4K] Joey’s scoring “in the Wth percentile” on the reading test means that Joey scored as well as or better than W'it of all students who tool: the reading test and scored worse than only 3%. His scoring in the End percentile on the mad-i portion ofthe test means that he scored as well as or better than "£2913 of all students who took the math test and worse than 23%. That is, Joey did better on the reading test, relative to his peers, than he did on due math test. 2.41 a} sssw. {b} asses. 4L... ”WEBER? III-w: 1:5! lus- :13! A five-1:.5Iizlil5 Tow: all: luv-=13! fire-1:552:95? Mu: 1'12 Inuit!!! {a} asses. flred=£EEFSF tutu-r El! lul‘=."l2 The Normal Distributions 55 2.42 {a} 98th percentile of Z = 2.854. {b} 2881 percentile of Z = .2222. 2.45 [2' = 25th percentile ofN{128,3[1} 149.265; Q; = 25th percentile of P411128, 38} = 198.235. Median = 58th percentile of 151(121}, 38] 1211'. ' - 2.44 {a} X s: 28 corresponds to Z s: [28 — 2512r 5 = — 1.81}. The relative frequency is .1582. Alternatively, normalcdft—lE‘J‘J, 28, 25, 5} = .1582. {b} X s: 10 corresponds to Z s: {11} — 252125 = —5.l1t1. The relative frequency is .8813, nonnalcdft—IE‘JGF, ll}, 25. 5} = .88135. {c} The top quarter correspunds to z = .625. Solving .825 = {x — 25}r’5 givea 28.38. 2.45 13. 2.46 The proportion scoring below 1.2 is about {1.852; the proportion between 1.2 and 2.1 is about [1.828. 2.42 Soldiers whose head circumference is outside the range 22.8 i 1.8], approximately, less than 21 inches or greater than 24.6 inches. 2.48 Those scoring at least 3.42 are in the “most FinglofEnglish" 58%; those scoring less than 2.58 make up the “most leleirican,‘r Spanish” 38%. 2.49 {a} Using the window dimensions shown, the histogram shows._a distribution that is fairly sym- metric with no obvious outliers [a boxplot shows that lfl.l2 is 'an outlier}. The mean {8.48) is approximately equal to the median {8.42}, an indication of symmetry. iltdxfl=1fl.12' [b] We remove 6.25 {the day after Thanksgiving], and the two largest times, 9.25 and 18.12 [icy roads and a delay clue to a traffic accident}. The remaining data are slightly skewed left; the mean {8.36} is less than the median {8.42}. 2.58 The normal probability plot for the weight gain for the chicks in the control group {normal corn] is: 36 Chapter 2 Iwhere the observed weight gain is on the xvaxis and the standardized value is on the y—axis. The normal probability plot for the experi mental group of chicks fed the lysine added diet is: Both plots show a fairly linear pattern of points, so WE conclude that both distributions are approx- imately normal. Thus it is reasonable to use I and s for the center and spread of the distributions. 2.51 {a} 0.2525. 25.25% of all children have scores greater than 110. Ht cu=.2 52w: 1aul=110 III-=1E55 {b} (1158?. l5.3?% of all children have scores lower than 35. firii=JfiflESE 1¢w= '1 E5! ul-=l5 {c} Using the original distribution Nflflfl, l5}: Calculate and display the area under the density curve of Nflflfl, 15} between TD and 131]. We obtain an area of [ll-7545. The Normal Distributions 5'! Using the 68—95-99} rule: By this rule. 95% of all observations of NUDE], 15} must fall within two standard deviations of the mean. The related area is (3.9599. 2.52 Exercise 2.23: normaicdf {—1E99. -2.25. El. 1] = 0.0122 normaledf 1-2. 25. lEQQ. I3, 1} = UHETS normalcdf (1.29. 11299, D. l} = 0.0334 normaledf (—2.25. 1.22. U. 1] '—- 0.9494 Exercise 2.41: normaiodf [—1E99 , 1.28. G. 13 = 13.399? normalcdf (—0.42. lEEJEt. CI. 1'; = 1115628 normalodf [—D.42. 1.28. . H = 1125:5215”~ CI riormalcdfi [—1E99. (1.42. CI. 1‘: | P on Ch. N on 2.53 No. 1|Within 4 standard deviations is .99995?‘ area. Going out to 5 standard deviations gives area .999999, which rounds to l for 4 decimal place accuracy. 2.54 Exerciso 2.2.9: invNorir. {{JJEI. El. 1} 0.52.4401'052 invNorm (0.15, U. 1] = -1.C|3154or —l.fl4 invNorm ii}. SCI. El. 1] —- [1.3415 or [1.84 invNorm [0.10. :J, l] = -1.282 or —l.28 Exercise 2.42: invNor-m [El . 98. CI. 1: = 2.054 or 2.05 invNorm [n.sa. G, 1.1 DETEZ or 9.7"? ...
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