Is Human Height Bimodal? Author(S): Mark F. Schilling, Ann E. Watkins, William Watkins Source: the American Statistician, Vol

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Is Human Height Bimodal? Author(s): Mark F. Schilling, Ann E. Watkins, William Watkins Source: The American Statistician, Vol. 56, No. 3, (Aug., 2002), pp. 223-229 Published by: American Statistical Association Stable URL: http://www.jstor.org/stable/3087302 Accessed: 12/04/2008 14:34

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Is Human Height Bimodal? Mark F. SCHILLING,Ann E. WATKINS,and William WATKINS

the distributionsof his male and female The combined distributionof heights of men and women has Although separate students are the of men and become the canonical illustrationof bimodalitywhen teaching approximatelynormal, histogram women is bimodal.Joiner "Note thatthis introductorystatistics. But is this example appropriate?This ar- together clearly wrote, has a bi-modal due to the of two ticle investigates the conditions under which a mixtureof two histogram shape mixing separate males and females."This idea in normal distributionsis bimodal. A simple justification is pre- groups, appealing appears several textbooks: sented that a mixtureof equally weighted normaldistributions introductory with common standarddeviation a is bimodal if and only if the differencebetween the means of the distributionsis greaterthan "Ahistogram of the heightsof studentsin a statisticsclass would be bimodal, 2a. More generally,a mixtureof two normaldistributions with for example, when the class contains a mix of men and women."(Iversen and similar cannot be bimodal unless their means differ Gergen 1997, p. 132) variability often occurs when data consists of observationsmade on two morethan the sum of theirstandard deviations. "Bimodality by approximately differentkinds of individualsor objects. For example, a histogramof heights of Examinationof nationalsurvey data on young adultsshows that college studentswould show one peak at a typical male height of roughly 70" the separationbetween the distributionsof men's and women's and anotherat a typical female height of about 65"." (Devore and Peck 1997, is not wide to We p. 43) heights enough producebimodality. suggest "If look at the of without out males and fe- reasons you heights people separating why histogramsof height neverthelessoften appearbi- males, you get a bimodaldistribution ..." (Wild and Seber 2000, p. 59) modal.

KEY WORDS: Bimodaldistribution; Living histogram;Nor- Still other textbooks include a problemthat asks studentsto mal distribution. predict the shape of the height distributionof a group of students when there are an equal number of males and females. 1. THE DISTRIBUTION OF HUMAN HEIGHT The expected answer is "bimodal."The distributionof height offers a plausibleexample of bimodalityand is easy for students Brian Joiner's (1975) living histogram(Figure 1) of his stu- to visualize. dents at Penn State the standard grouped by height inspired Ratherthan sending our studentsout onto the football field a classroom of from a mixture of example bimodality resulting la Joiner to demonstratebimodality, we decided to get some two populations.

Figure 1. Joiner'sliving histogram of student height.

governmentdata and construct the approximatetheoretical density function for the mixture of the male and female populations. The most recent National Health and Nutrition Exami- MarkF. Schillingis Professor,Ann E. Watkinsis Professor,and William nation Survey (NHANES III), conducted in 1988-1994 by the Watkinsis Professor,Department of Mathematics,California State University, NorthridgeNorthridge, CA 91330-8313(E-mail: [email protected]). The United States National Center for Health Statistics, reportsthe authorsthank Rebecca Walkerand DeborahNolan for helpful comments. cumulative distribution of height in inches for males and for fe-

( 2002 AmericanStatistical Association DOI: 10.1198/00031300265 TheAmerican Statistician, August 2002, Vol.56, No. 3 223 males in the 20-29 age bracket (U. S. Census Bureau 1999). of the other distribution(Figure 3b), then outside the compo- The datafor each sex have the means and standarddeviations in nent modes the mixtureis monotone, while between modes the Table 1 and each follow a normaldistribution reasonably well. mixtureis the sum of two linearfunctions, hence is itself linear. Thus the mixtureis not bimodal. Table1. SummaryStatistics for NHANES Height Data Now considermixture densities of the form f(x) = pfl (x) + Mean SD* (1 -p)f2(x), where fl and f2 are each normaldensities with means < and variances o2 and and 0 Males 69.3 2.92 p1 12 -2,respectively, < Females 64.1 2.75 p < 1. Considerfirst the simplestsituation where a2 = oj2 = a2 and p = 0.5. If I1t and I2 are far apart,then f clearly will be *Standard deviations were computed from the cumulative distributions, as they were not bimodal,resembling two bell curves side by side. This certainly supplied by NHANES. happens when 12 - P1 is greater than about 6a. Furthermore, it is easy to see that the modes of f occur not at 1I and I2, but between them: Both and are Using the NHANES means and standarddeviations as param- fl f2 strictlyincreasing below pi, so has derivativethere. At eters for two normal densities and assuming equal numbersof f positive p1i, fi has a derivativeof zero and each sex, we get the graphin Figure2 for the theoreticalmixture f2 is increasing,so f has positive derivativeat 111.Thus mode must be distributionof height for persons aged 20-29. any largerthan /t1. Similarly,any mode must be smaller than 12- But what happenswhen I,1 and I2 are much closer together? The result for this case is generally creditedto Cohen's (1956) problem in the American Mathematical Monthly, but dates back at least to Helguerro(1904):

Theorem 1. (Helguerro 1904). Let fl and f2 be normal densities with respective means It1 and 12 and common variance a2, and let f be the mixturedensity 0.5 fi + 0.5 f2. Then f is unimodal if and only if I12 - 1t l < 2ca. Helguerroand the Monthlyinclude a proof of the seemingly obvious fact that a mixture of two normal densities must be either unimodal or bimodal. With that assumption, a simpler 55 60 65 70 75 80 justificationof Theorem 1 appearsbelow. Figure2. Theoreticaldistribution of U.S. young adulthuman height in A normal is inches as a mixtureof twonormal distributions using means andstandard Proof: density concave down between its inflec- deviationsfrom NHANES data. tion points Iu? a and concave up elsewhere. Note that f is symmetric around m = (1I + 1U2)/2. Now if 112 - 11I > 2a, This obviously is not a bimodal distribution!Yet Joiner'sliv- thenboth fl and f2 areconcave up at m, hence so is f. Therefore f has a local minimumat which that is ing histogramshows two clear peaks. We startour investigation m, implies f bimodal. if - of this paradox by studying the modality of a mixture of two Conversely, I12 lI < 2c, then both fl and f2 are concave down at hence so unimodaldensities, with particularfocus on normaldensities. m, is f. Thus, f has a local maximumat m and is thereforeunimodal. In the borderlinecase 1t2 - 11 = 2a the second derivativeof f vanishes, but the fourthderivative can be 2. AN INVESTIGATION OF BIMODALITY used to show that f has a maximumat m and thus is unimodal. Most students are willing to believe that the mixture of two unimodaldensities with differingmodes will necessarilybe bimodal, as each of the componentmodes will generatea peak in the mixture distribution.It is quite easy to show that this is not a b the case. Consider a mixture of two triangulardistributions, as shown in Figure 3. If neitherdistribution's the - - supportoverlaps -Imb.- - am, -,__ m m Figure 4. (a) When pi1 - I|21<2a, both normal components (light lines) are concave down at m = (i 1 + IL2)/2, so the mixture f (heavy line) is also concave down at m and is therefore unimodal. (b) When Il1 - 12 1>2a, both components are concave up at m, so the mixture f is also concave up at m, hence is bimodal. The dots indicate inflection 3. Dashed Figure lines represent an equal mixture of the component points of the component densities. distributions (solid lines). Figure 4 illustratesthis argumentthat a two standarddevi- other'smode then the mixturedistribution is (Figure3a), indeed ation separationbetween the means is needed for bimodality. bimodal. if each mode is However, containedwithin the support The symmetryin Figure 4 is crucial and so this proof does not 224 Teacher'sCorner Table2. The MixtureDensity f = pf1 + ( 1 - p)f2 Is BimodalIf and Only Figure 5 shows that S(r) is close to 1 for cases in which the - If 1P2 11\ Exceeds (af1 + 02) Times the Value Indicated component variances are at all similar (as with heights), thus giving us the following rule of thumb: oai/2 p= 0.3 p= 0.4 p= 0.5 p= 0.6 p= 0.7 Regardless of the mixture proportion p, a mixture of two nor- 1.00 1.36 1.22 1.00 1.22 1.36 mal densities with approximately equal variances cannot be bi- 0.95 1.35 1.20 1.08 1.24 1.36 modal the in means is much less than the sum 0.9 1.34 1.16 1.11 1.25 1.36 if separation of 0.8 1.29 1.01 1.15 1.26 1.35 the component standard deviations. 0.7 1.18 1.06 1.16 1.24 1.32 Although Figure 5 indicates that bimodality occurs for cer- 0.6 0.96 1.06 1.13 1.20 1.26 tain mixtureproportions if the separationin means is less than 0.5 0.96 1.03 1.08 1.14 1.19 or equal to al + 02, for most mixtureproportions a somewhat greaterseparation is required.Table 2 gives the factorfor al +7a2 that unimodaland bimodal mixturesof normaldensi- generalizeto the case when V=o-2 oV2,as with the heights of men separates and women, or when there is an unequalmixture of fl and f2, ties for specific values of the mixtureproportion p and various = which will occur more often thannot in classroomexperiments. choices of C0l/r2. For example, if 1/02 0.9 and p = 0.6, The modality of an arbitrarymixture of two normal den- then the mixtureis bimodal if the means are fartherapart than sities involves a complex interplay between the difference in 1.25(al + 0-2); if not, it is unimodal. means, the ratio of the variances, and the mixture propor- Mixtures that are bimodal when the separationin means is tions. Strong conditions for unimodality/bimodalitywere de- less than the sum of the standarddeviations have only a slight rived by Eisenberger(1964) and Behboodian (1970). Robert- dip if the standarddeviations are at all similar.Figure 6 shows son and Fryer (1969; reproducedby Titterington,Smith, and the mixture with the most pronounceddip possible among all Makov 1985) obtainedprecise specificationsof when bimodal- mixturesof two normaldensities in which 0.6 < o1l/02 < 1. ity occurs. (See also Kakiuchi(1981) and Kemperman(1991), who addressedthe question withoutthe normalityassumption.) We have attainedbimodality conditions equivalentto those of Robertsonand Fryer,although we deriveand presentthem quite differently.These results follow from the observationthat in order for a mixture f(x) of normal densities to be bimodal there must be at least one value x where f(x) satisfies both f'(x) = 0 and f"(x) > 0, so that f(x) possesses a relative minimum. Solving pf(x) + (1 - p)f2(x) 0 and pf '(x) +(1 -p)f2 (x) > 0 simultaneously yields the inequality f2 (x)/f (x) < f2 (x)/ft(x). After extensive algebraic manip- Figure6. Mixtureof twonormal densities witha1 = 0.6022, I12--L 1 = ulationsof this inequality,we obtainthe following generalization a1 + 02, = .288. This has the most pronounceddip of any of Theorem 1. p density mixture in which 0.6 < U1/c2 < 1 and 12 - 11i <_ +1 + 2- Let r = o-2/U211 22al and define~ll theIL ttUUseparation CIJLV factor S(r)J\ toVV be

- -2 + 3r+3r2 2r3 + 2(1 - r + r2)2 S(r)= v(1 + Vr) 3. THEORY VS. DATA Thenthe mixturedensity f pfl + (1 -p) f2 is unimodalfor all the theoreticaland numericalresults of Section 2 give p if and only if 1/2- 1 | < S(r)( l + U2). Note that S(1) = 1, Although some for when a mixture be unimodal which for the case p = 0.5 yields Theorem 1. guidelines density might or bimodal, the modality issue becomes more complex when S one is dealing with actual data such as students' heights. The 1.00 study of bimodality and multimodalityfor data has a long and extensive history,beginning with Pearson (1894). Many inves- tigators have developed methods for resolving the distribution 0.75 into its underlying(typically Gaussian)components, as well as testing whetherthe numberof these componentsis greaterthan 0.50 one. See Holgerssonand Jomer (1978), Everittand Hand (1981), and Titterington,Smith, and Makov (1985) for good overviews 0.25 of the subjectand extensive references. This section focuses on the modality of human height as it appearsin sample data. In particular,why is Joiner'sliving his- 1 2 4r togram bimodal? If we look at the NHANES data of female Figure 5. Separation factor S(r) needed for possible bimodality of a and male heights, al/0r2 = 2.75/2.92 = 0.94. According to mixtureof two normaldensities, where r is the ratioof theirvariances. If the second row of Table 2, a separationin means of at least \L2 - A11 < S(r)(a-1 + (2), any mixture proportion results in unimodality. 1.08(2.75 + 2.92) = 6.1 inches is necessary for bimodalityof the correspondingmixture distributionfor most typical class-

TheAmerican Statistician, August 2002, Vol.56, No. 3 225 Figure7. Livinghistogram of 143 studentheights at Universityof Connecticut. room proportionsof males and females. But the difference of darddeviations as the Strausbaughsample and the same mixture 5.6 inches in the men's and women's means in Joiner's sample proportion-it could easily pass for a bell curve, although the (Figure 1) is less than that. top is slightly less peaked. Yet Joiner is not the only professor to get a bimodal distribution when male and female heights are combined.The living histogramin Figure7, which appearedin TheHartford Courant (1996), Crow(1997), andWild andSeber (2000), shows the male andfemale genetics studentsof LindaStrausbaugh at the Univer- sity of Connecticutin 1996. Crow writes, "Since both sexes are included, the distributionis bimodal."The effect is somewhat bimodal, but only slightly. The StatLab (Hodges, Krech, and Crutchfield1975) data of 1296 male and 1296 female heights in the San Francisco area also somewhatbimodal. Bay appear 55 60 65 70 75 80 We now considerpossible reasonsfor this discrepancybetween Figure8. Mixtureof twonormal densities withthe same means, stan- data and theory. darddeviations, and mixtureproportions as the Strausbaughsample. 3.1 Bad Parameters We cannotsay for surethat the NHANES means and standard Indeed,unimodality is quite robust:as the means of the com- deviations in Table 1 apply precisely to college students. The ponent normaldensities separate,any bimodalityremains quite NHANES sample consists only of around600 of each sex aged unremarkableat first (see Figure 9). For an equal mixture of 20-29 andthe heights are self-reported.College studentsare not two normal densities having the same standarddeviation a to demographicallyidentical to the generalpopulation aged 20-29. have a of 36%, like that in the Joiner a And Joiner'shistogram was made 20 years before the NHANES dip histogram,requires study. separationin means of about 3a. However, theoretical mixtures that use the sample means, standarddeviations, and mixture proportionsfrom the Joiner and Strausbaughsamples as parametersare still not bimodal. For Joiner's photograph,we found means of 64.6 inches for females and 70.2 inches for males, and respective standarddeviations of 2.6 and 2.8 inches. The group is 54% female. (We are uncertainabout the exact counts for some heights but not enough to affect the analysis significantly.)The difference in means is 5.6 inches, which slightly exceeds the sum of the sd's, 5.4 inches, but the separationneeded for bimodalityin this case is about 1.12(ai + J2) = 6.05 inches. The computed means and standarddeviations in inches for Strausbaugh'sstudents are 64.8 and 2.7 for females and 70.1 -4 -2 2 4 and 3.0 for males; the mixture is about 45% female. Here the Figure9. Six equal mixtures(p = .5) of two normaldensities. Each differencein means is not even as large as the sum of the sd's, componenthas a standarddeviation of 1. The values of 1\/2 - l11 are and is again too small for bimodality.Figure 8 shows a mix- 2.0, 2.2, 2.4, 2.6, 2.8, and 3.0. tureof two normaldistributions with the same means and stan-

226 Teacher'sCorner 3.2 Bad Model 3.3 Bad Data The normaldistribution is only an approximationto the dis- We cannot be confident that the students lined up at their tributionof human height. What if the true distributionwithin actualheights. For example, Figure 10 shows the heights of the each sex is somewhat different?Kemperman (1991) provided men in a class at the University of Californiaat Davis (Saville necessary and sufficient conditions for a mixture of unimodal and Wood 1996). The instructorsent a sheet of paper around componentdensities satisfying certainweak prerequisitesto be the class and asked studentsto record their height and gender. itself unimodal.These constraintsare rathercomplex; however The spike at six feet cannot reasonablybe attributedto chance the concavity argumentsof Theorem 1 lead easily to a simple variation,and it looks very much like students an inch or two conditionfor the possibility of a bimodalmixture when normal- shorterrounded their heights up to six feet. ity is not assumed: Separatingthe Universityof Connecticutstudents by gender shows thatthe impressionof bimodalityis largely due to a spike Theorem2. Let fl and f2 be continuousand differentiable in women's heights at 5' 6" and spikes in men's heights at 5' 10" densitieson the real line. Let fl be decreasingand concave up for and six feet, apparentlypopular heights when lining up with x > P, andlet f2 be increasingand concave up for x < Q, where classmates. The Joiner sample also has a modest peak at six P < Q. Thenthere exist mixturesf(x) = pfi (x)+(1l-p) f2 (x) feet. that are not unimodal. More subtly,students must round their height to inches. If men like to be thoughtof as tallerand women preferto be thoughtof Proof: Let x be any point in the interval (P, Q). Since at x, as shorter(more likely for the women in 1975 than today), then it men roundtheir to fl is decreasingand f2 is increasing,it is possible to choose p would be reasonablefor the to heights up inch and for the to down to the next inch. so that f'(x) - pf (x) + (1 -p)f2(x) 0. In addition, f is the next women round That would an extra 1" between the means concave up in (P, Q) since fl and f2 are. Thus f has a local produce separation minimumat x, hence is not unimodal. comparedto roundingto the nearestinch. The resultingmixture is bimodal, althoughthe dip is so slight (just 0.16% below the For example, it is known that there are somewhatmore very lower relative maximum) as to be nearly undetectable.Figure tall andvery shortpeople thanwhat would be foundfor a perfect 11 is the counterpartof Figure 2 but with an extra one-inch normaldistribution, and so a properlyscaled t distributionmight separationin the means. be a reasonablealternative model. Student'st distributionwith v degreesof freedomhas inflectionpoints at ? V/2v/(2v + 1). An equal mixtureof two t distributionsshifted apartby more than 2 /2v/(2v + 1) thereforesatisfies the conditionsof Theorem2 with its local minimumoccurring at the center of symmetryof f, and the mixture is bimodal. For the t distributionwith v 10, the required separationis 2/2 .10/(2 -10 + 1) = 1.95. This distributionhas a standarddeviation of /v/(v - 2) = 1.12. The distributionsof the heights of males and of females have standarddeviations of approximately2.8 inches. Thus the means need to be separatedby (1.95)(2.8/1.12) = 4.9 inches, less than what is requiredwith the normal model. The mean differencein the NHANES datais 5.2 inches, so with the t model, 11. Theoreticaldistribution of U.S. adulthuman bimodality would result, though the dip would be very small. Figure young height, if males round to the next inch and females rounddown. The availablepercentiles from NHANES, however, actually fit up the normalmodel betterthan a t model with small to moderate of freedom. degrees 3.4 Bad Bins Relatedto the issue of how studentsround their heights is the Frequency broaderissue of placing continuous data into bins. When using a histogramto determinewhether a set of data has multiple 15- modes, certainbin widths can hide the bimodal structureof the data.For example, Wand(1997) showed how the S-Plus default 10- bandwidthoversmooths a histogramof Britishincomes. But we have the opposite problem-the smooth, mound-shapeddistri- butionof Figure 1 being placedinto one-inchbins andapparently coming out bimodal. The width of the bins cannot explain that. ^ lli 7 I 64 68 72 76 3.5 Variability Due to Sampling Height in Inches Figure 10. Self-reportedheights of 77 male students at U.C.Davis. The earliest living histogram that we can find (Figure 12) shows the heights of studentsat ConnecticutState Agricultural

The AmericanStatistician, August 2002, Vol.56, No. 3 227 4:10 4:11 5:0 5:1 5:2 5:3 5:4 5:5 5:6 5:7 5:8 5:9 5:10 5:11 6:0 6:1 6:2 Figure 12. Livinghistogram of 175 male college students (Blakeslee 1914).

College (now the University of Connecticut).It looks bimodal ball field and found themselves with nothinginteresting to pho- too, with modal heights at 5' 8" and 5' 1"-but these students tograph. were all men! The explanation is not bad parameters,a bad model, or bad data: "Each of the 175 students shown ... was 4. SUMMARY measuredin his stocking feet and placed in the rank to which A mixtureof two normaldensities will not be bimodalunless his height most nearly corresponded"(Blakeslee 1914). What there is a difference between their could explain this? very large means, typically than the sum of their standarddeviations. If male and The distributionof bin counts in a histogramis a multinomial larger female areeven distributedwith randomvariable with cell probabilitiesdetermined from the un- heights approximatelynormally means and standarddeviations close to those the derlying distribution.There is more variabilityin the bins that reportedby NHANES the difference between the means of male have the largercounts than in the bins that have smallercounts. survey, and female is not to Yet A density that is the sum of two normal densities with nearly heights enough produce bimodality. of do show equal standarddeviations and means that are almost two stan- photographs living histograms bimodality. We have discussed several reasons a of male darddeviations apart is mound-shapedwith a relativelyflat top. why histogram andfemale have a bimodal even In the case of an equal mixture of two normal densities with heights may appearance though, reasonable estimates of the the parametersmatching the NHANES data, only six bins (64" to using parameters,theory says distributionis For the ten- 69") hold the middle 53% of the heights and their relative fre- underlying mound-shaped. example, of students to favor certain values when quenciesare quite similar:0.085, 0.092, 0.093, 0.092, 0.088, and dency self-reporting their contributeto the And when tak- 0.083. Withthis relativelyconstant central region, small number height may phenomenon. small random from a it of bins, andhigh variabilityin theircounts, it will not be unusual ing samples mound-shapeddistribution, is common to a that looks bimodal. for samples to give two nonadjacentbins that appeartaller than get histogram the others. So what'san instructorto do if he or she decides tojettison this We haven't been able to find an The destructionof smoothness by sampling is well known. compelling example? equally classroomdemonstration of as the mixture Murphy (1964) constructed 12 histograms for samples of 50 appealing bimodality of two randomnormal deviates. All but about four have enough irreg- populations. We ran some simulations to see what conditions would be ularityto be reasonablycalled bimodal.We generateda number to consistent of true in a of simulated histogramsof heights ourselves; very few had a likely give representations bimodality classroom With standarddeviations and sam- clear unimodalappearance. The literatureestablishes that using experiment. equal sizes of 25 or more a in the means of the multimodalityof a histogramto infer multimodalityof the ple per group, separation about three standarddeviations results underlyingdistribution is unreliableunless the sample size is usually produces good when around15-20 intervals. the very large. Both false positive and false negative errorsoccur using Occasionally histogram is too to be identified as and in rare frequently. irregular clearly bimodal, instanceslooks more unimodalthan bimodal. Such cases are of course more rare for larger sample sizes than for small ones. 3.6 Selection Bias Perhapsa readerwill be able to provide an easy-to-implement Although the living histogramof heights is often used as an classroom demonstrationhaving approximatelythis degree of example of a bimodal situation,we have seen only two actual separationbetween the components. photographsthat purport to show this bimodality.One can only wonderhow many instructorsgot their studentsout on the foot- [Received May 2001. RevisedAugust 2001.]

228 Teacher'sCorner REFERENCES 339-340. Kakiuchi,I. (1981), "UnimodalityConditions of the Distributionof a Mixtureof J. "On the Modes of a Mixture of Two Normal Distribu- Behboodian, (1970), Two Distributions,"Kobe University.Mathematics SeminarNotes 9,315-325. tions,"Technometrics, 12, 131-139. Kemperman,J. H. B. (1991), "Mixtureswith a Limited Number of Modal In- Blakeslee, A. F. "Cornand Men,"The Journal 5, 511-518. (1914), of Heredity, tervals,"The Annals of Statistics, 19, 2120-2144. Cohen, A. C. Normal Distribution" Problems (1956), "Compound (Advanced E. A. (1964), "One Cause? Many Causes? The Argumentfrom the and AmericanMathematical 129. Murphy, Solutions), Monthly,63, Bimodal Distribution,"Journal of ChronicDiseases, 17, 301-324. Crow, J. F. "BirthDefects, Jimson Weeds and Bell Curves,"Genetics, (1997), Pearson, K. (1894), "Contributionto the MathematicalTheory of Evolution," 147, 1-6. Philosophical Transactionsof the Royal StatisticalSociety of London,Ser. A, Devore, J., and Peck, R. (1997), Statistics: The Exploration and Analysis of 185,71-110. Data, Belmont, CA: Duxbury. Robertson,C. A., and Fryer,J.G. (1969), "Some DescriptiveProperties of Nor- I. "Genesis of Bimodal Eisenberger, (1964), Distributions,"Technometrics, 6, mal Mixtures,"Skandinavian Aktuarietidskrift, 52, 137-146. 357-364. Saville, D. J., andWood, G. R. (1996), StatisticalMethods: A GeometricPrimer, B. and D. J. Finite Mixture London: Everitt, S., Hand, (1981), Distributions, New York:Springer, 66. and Hall. Chapman Titterington,D. M., Smith,A. F.M., andMakov, U. E. (1985), StatisticalAnalysis The Courant New November Hartford (1996), "Reaching Heights," 23, 1996; of Finite MixtureDistributions, Chichester: Wiley. K. photo by Hanley. U.S. CensusBureau (1999), StatisticalAbstract of the UnitedStates: 1999. Table F. "Sui Massimi Delle Curve Helguerro, (1904), Dimorfiche,"Biometrika, 3, #243, p. 155. Source of table: U.S. National Centerfor Health Statistics,un- 85-98. publisheddata. Data apparentlyreorganized from: U.S. Departmentof Health Hodges, J. L. Jr.,Krech, D., andCrutchfield, R. S. (1975), StatLab:An Empirical andHuman Services (DHHS), NationalCenter for HealthStatistics, Third Na- Introductionto Statistics, New York:McGraw-Hill. tional Health and NutritionExamination Survey, 1988-1994, NHANES III Holgersson, M., and Jorner,U. (1978), "Decompositionof a Mixtureinto Two LaboratoryData File. Public Use Data File DocumentationNumber 76200. NormalComponents: A Review,"International Journal ofBio-Medical Com- Hyattsville,MD: Centersfor Disease Controland Prevention,1996. puting, 9, 367-392. Wand.M. P. (1997), "Data-BasedChoice of HistogramBin Width,"The Amer- Iversen, G. R., and Gergen, M. (1997), Statistics, the ConceptualApproach, ican Statistician,51, 59-64. New York:Springer. Wild, C. J., and Seber, G. A. F. (2000), Chance Encounters:A First Course in Joiner,B. L. (1975), "Living Histograms,"International Statistical Review, 3, Data Analysis and Inference,New York:Wiley, pp. 58-60.

TheAmerican Statistician, August 2002, Vol.56, No. 3 229