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}}}{EXCHG {PARA 206 "" 0 "" {TEXT 200 36 "Densities and Distribution Functions" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 298 "IMPORTANT NOTE! In this no tebook we will use a procedure to draw histograms that corresponds m ore closely to the presentation in the text, as well as some simulate d data we will use in our examples. Rather than clutter the worksheet with these, we have placed them in a hideable subsection. " }{TEXT 206 111 "In order to activate the procedures and make the data availab le for your Maple session, first do the following:" }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 5 "" 0 "" {TEXT 227 234 "Click on the + (right-pointing triang le in Maple 10) and execute the commands by clicking again on the firs t command line. It is sufficient to click on ONLY the FIRST line. The n click - (down-pointing triangle in Maple 10) to hide." }}{PARA 0 " " 0 "" {TEXT 226 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "with (stats):with(stats[statplots]):with(random):with(statevalf):\n" } {MPLTEXT 1 0 14 "with(plots):\n" }{TEXT 226 118 "The following procedu re improves the histogram program. It allows specification of bin ra nges along with thedata. " }{MPLTEXT 1 0 2 "\n" }{MPLTEXT 1 0 55 "hi sto:=proc(data,ranges) local n,i ; n:=nops(ranges);\n" }{MPLTEXT 1 0 55 " histogram(stats[transform,scaleweight[1/nops(data)]]\n" }{MPLTEXT 1 0 39 "(stats[transform,tallyinto['extras']]\n" }{MPLTEXT 1 0 57 " \+ (data,[seq(ranges[i]..ranges[i+1],i=1..(n-1))]))): end:" }{TEXT 226 30 "Here is the simulated data: \n" }{MPLTEXT 1 0 1396 "normaldataset \+ := [.8441162075, .1675078072, -.6676777329, -.4995752138e-1, 1.5021551 18, -.5568011924, 1.707902213, -1.074078825, .4199611949, -.9577805554 e-2, .1697413106e-1, -.4995798757, -.6337893330, -.5650013958, .174413 6103, -.6279057848, .8220447994, .6040674919, .6442369504, -1.32760138 6, 1.669060099, 1.013599091, -1.206729633, -.9813553812, -1.227766719, -.6001401143, -.3683659020, .1020625770e-2, -.5613021077, -1.23776757 2, .5907551397, .1688781208e-1, -.1614290579, -.2631929959, -.26828269 64, .9013466929, -.1423211060, 1.575322000, 1.968681058, -.7095172514, .2476207045, -.1310654436, -.1627628270, -.9528732852, .2177892855e-1 , .2941735009, -.2621070194, -1.670050594, .8984387349, 1.063402533, . 1685053465, 1.233355953, .4445350600, 2.030521026, -1.291595763, -2.61 8596553, 2.035686170, -.2022766430, .1000936624, .5039054945, -.409571 4084, .2492422437e-1, -1.050241620, -1.532729703, -.4053617765, -.2600 474204, 1.743665944, 1.421809036, -.5183661918, -1.261056576, -.871950 9654, -.2055324937, 1.269669973, .7711251851, -2.356327134, .496413577 2, -1.159187869, -.2856297104, .1910937487, .7958866242, .6278507547, \+ -.7620564877, .2706999082, -.7669122519, .5143784655, -.3783585777, -. 3311418580, -.8202443397, .3956496264, .9380667145, .7823346678, 1.226 855457, .4931329422, -1.673055015, .4068140517, -.8048465492, -.282277 6423, .9585722598, -.4911639986, -.2011264576e-1]:\n" }{MPLTEXT 1 0 1398 "expdataset := [7.885590925, 4.702060218, 2.490185865, 3.24224065 3, 1.340675124, 3.661654543, .8143198115, .3448392635, 1.762374708, 1. 596616308, 3.045972928, .1048340351, .8122487665, 1.312042270, .321070 8275, .6063731865, 8.025586013, .3103939710, .4247798420, .7083070305, 2.557072468, 11.75270413, 5.613973343, 1.030900203, 3.763702180, .835 9464300e-1, 2.840576003, 1.431463221, 1.451136681, 5.397233638, 3.8718 91675, 3.712009483, 2.366286344, .1507242052, 1.529106515, 4.819369353 , .6740114550, 4.372344810, .2772995553, 8.880719618, 1.559404730, .23 09232676, .4691102838, 3.527392048, 2.034073253, 4.462702103, 2.176072 523, .6497600298, .1098657740, 4.683669508, 2.383357198, .4602255753, \+ 2.876770555, .2158049156, 3.315624530, 2.321797189, .3351986210, .7641 193595, 8.389897488, 1.712383348, 2.294743645, .2008303222, .734998926 0, .9877686680, 3.034921273, 2.246621681, .6304389260, .8058789315, .6 728238965, 1.042509567, 5.434997463, .8238934258, 4.493314183, .466463 4878, 5.888566448, 2.228198160, 4.002202828, 1.348413776, 1.153161079, .1611020836, 2.339653543, 2.711198408, .2406473455e-1, 4.704403665, . 1179580323e-1, 3.091762885, .1785056384e-2, 1.861998363, 1.958990347, \+ .7647498905e-1, 5.275310273, 1.719625418, 9.575994495, 2.764027188, .2 264810046, 5.890674828, .9424533133, .6675375813, 1.558428403, .250615 9995]: data1:=[seq(1.5*normaldataset[i]+12, i=1..nops(normaldataset))] : \n" }{MPLTEXT 1 0 42 "data2:=[seq(0.5*(data1[i]-4),i=11..90)]:\n" } {MPLTEXT 1 0 992 "expdata := [.1120369893, 2.334139781, 4.678167130, . 4705372194, 2.328780876, 12.34825991, 9.014209867, 5.436187016, 9.9137 48387, 1.540165607, 19.76166247, 4.701175160, .9907759420, 2.174687245 , 16.78005340, 45.34618402, 1.535887240, .4824757234, 14.30887167, 6.4 63480786, 17.40341021, 16.93452753, 4.394551810, 3.732709666, 23.10634 982, 22.25558379, 2.659917653, 13.93220691, 19.25247833, .8331147647, \+ 3.710851430, .6267695523, .6518215197, .9739978074e-1, 4.576383582, 21 .61790771, 4.504740148, 29.06304041, 2.020316093, 3.631875673, 19.8723 5464, .5335717365, 1.566086725, 1.963076922, 6.959548880, 2.490692867, .2907234955, 8.316983454, 16.80554524, 4.334713517, 2.205953864, 13.5 0022949, 6.937006914, 7.468602780, 4.926504421, 5.453933197, .14360843 29, 25.36055519, 3.686491281, 3.961540008, 14.60758545, 16.16638024, 1 1.23419654, 2.563366288, .2886807971, 4.362667264, 3.398720113, 9.7020 74154, 6.361943260, 4.944042345, 26.19876319, .5792690188, 10.60850810 , 8.331003234, 21.06726548]:\n" }{MPLTEXT 1 0 16 " data3:=expdata:" }} {PARA 0 "" 0 "" {TEXT 226 92 "SCROLL BACK TO TOP AND CLICK ON - (dow n-pointing triangle in Maple 10) TO HIDE ALL THIS!" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 148 "As usual, you should execute all the co mmands below as you read through the worksheet and before working on h omework problems in your new worksheet." }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}}{EXCHG {PARA 207 "" 1 "" {TEXT 200 45 "Describing data with hi stograms and densities" }}{PARA 0 "" 0 "" {TEXT 226 5 " " }}{PARA 0 "" 0 "" {TEXT 207 10 "Example 1." }{TEXT 226 106 " In an effort t o understand the weight variation of a certain species of small animal , a biologist has " }}{PARA 0 "" 0 "" {TEXT 226 246 "collected laborat ory data on 100 animals raised in a controlled environment. She belie ves that this is a representative sample of the species. The data is collected in a Maple list, giving weight in kilograms, which we call \+ data1. Here it is:" }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "data1;" }}{PARA 208 "" 1 "" {XPPMATH 20 "6 #7`q$\"+JuhE8!\")$\"+rh7D7F&$\"+S$[)*4\"F&$\"+sj]#>\"F&$\"+oKKD9F&$\"+ @)zk6\"F&$\"+K`=c9F&$\"+w\")))Q5F&$\"+zT*HE\"F&$\"+HLc)>\"F&$\"+?ha-7F &$\"+>I1D6F&$\"++;$\\5\"F&$\"+\"z\\_6\"F&$\"+U?;E7F&$\"+KT\"e5\"F&$\"+ ?nIB8F&$\"+C,h!H\"F&$\"+Vbj'H\"F&$\"+#zf3+\"F&$\"+:!f.X\"F&$\"+k)R?N\" F&$\"+b0**=5F&$\"+$p'z_5F&$\"+#*\\$e,\"F&$\"+$)*y*46F&$\"+:^uW6F&$\"+% 4`,?\"F&$\"+%o/e6\"F&$\"+k[L95F&$\"+rKh)G\"F&$\"+sJ`-7F&$\"+Tcyv6F&$\" +^5_g6F&$\"+'fd(f6F&$\"+/??N8F&$\"+M=ly6F&$\"++$)HO9F&$\"+f@I&\\\"F&$ \"+7Cd$4\"F&$\"+1J9P7F&$\"+$=S.=\"F&$\"+wbev6F&$\"+2!pq0\"F&$\"+RoE.7F &$\"+Dg7W7F&$\"+ZRog6F&$\"+4T#\\\\*!\"*$\"+5ewM8F&$\"+!Q5&f8F&$\"+-eFD 7F&$\"+$R.]Q\"F&$\"+f-om7F&$\"+a\"yX]\"F&$\"+O1E15F&$\"+q^5s!)F_q$\"+E HN0:F&$\"+/&e'p6F&$\"+\\S,:7F&$\"+Ceev7F&$\"+*Gk&Q6F&$\"+M'QP?\"F&$\"+ dPYU5F&$\"+Xa!4q*F_q$\"+Md>R6F&$\"+()G*4;\"F&$\"+#*)\\:Y\"F&$\"+b8F89F &$\"+r]CA6F&$\"+9:%3,\"F&$\"+bt?p5F&$\"+E, 8F&$\"+8w<%H\"F&$\"+F:p&3\"F&$\"+')\\gS7F&$\"+iJ'\\3\"F&$\"+qn:x7F&$\" +8iCV6F&$\"+@(G.:\"F&$\"+\\L'p2\"F&$\"+WuMf7F&$\"+2+rS8F&$\"++-N<8F&$ \"+>$GSQ\"F&$\"+T*pRF\"F&$\"+xuT!\\*F_q$\"+3@-h7F&$\"+=IFz5F&$\"+a$ew: \"F&$\"+ReyV8F&$\"++aKE6F&$\"+.J)p>\"F&" }}}{PARA 209 "" 0 "" {TEXT 200 2 " " }{TEXT 200 92 "In order to get a better feeling for the da ta, we first sort it using Maple's sort command:" }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "sort(data1) ;" }}{PARA 210 "" 1 "" {XPPMATH 20 "6#7`q$\"+q^5s!)!\"*$\"+*H4bY)F&$\" +xuT!\\*F&$\"+4T#\\\\*F&$\"+Xa!4q*F&$\"+#zf3+\"!\")$\"+O1E15F1$\"+9:%3 ,\"F1$\"+k[L95F1$\"+#*\\$e,\"F1$\"+b0**=5F1$\"+?=7E5F1$\"+w\")))Q5F1$ \"+dPYU5F1$\"+$p'z_5F1$\"+2!pq0\"F1$\"+bt?p5F1$\"+\\L'p2\"F1$\"+=IFz5F 1$\"+iJ'\\3\"F1$\"+F:p&3\"F1$\"+7Cd$4\"F1$\"+S$[)*4\"F1$\"++;$\\5\"F1$ \"+KT\"e5\"F1$\"+$)*y*46F1$\"+\"z\\_6\"F1$\"+%o/e6\"F1$\"+@)zk6\"F1$\" +r]CA6F1$\"+>I1D6F1$\"++aKE6F1$\"+*Gk&Q6F1$\"+Md>R6F1$\"+8iCV6F1$\"+:^ uW6F1$\"+@(G.:\"F1$\"+Vb:d6F1$\"+a$ew:\"F1$\"+'fd(f6F1$\"+^5_g6F1$\"+Z Rog6F1$\"+()G*4;\"F1$\"+E,\"F1$\"+.J)p>\"F1$\"+HLc)>\"F1$\"+%4`,?\"F 1$\"+sJ`-7F1$\"+?ha-7F1$\"+RoE.7F1$\"+M'QP?\"F1$\"+\\S,:7F1$\"+rh7D7F1 $\"+-eFD7F1$\"+U?;E7F1$\"+iSmG7F1$\"+1J9P7F1$\"+')\\gS7F1$\"+Dg7W7F1$ \"+WuMf7F1$\"+3@-h7F1$\"+zT*HE\"F1$\"+f-om7F1$\"+T*pRF\"F1$\"+P?Yu7F1$ \"+Ceev7F1$\"+qn:x7F1$\"+rKh)G\"F1$\"+C,h!H\"F1$\"+8w<%H\"F1$\"+Vbj'H \"F1$\"+y(ocJ\"F1$\"++-N<8F1$\"+%*HQ>8F1$\"+?nIB8F1$\"+JuhE8F1$\"+5ewM 8F1$\"+/??N8F1$\"+2+rS8F1$\"+ReyV8F1$\"+k)R?N\"F1$\"+!Q5&f8F1$\"+>$GSQ \"F1$\"+$R.]Q\"F1$\"+'\\]/R\"F1$\"+b8F89F1$\"+oKKD9F1$\"++$)HO9F1$\"+: !f.X\"F1$\"+K`=c9F1$\"+#*)\\:Y\"F1$\"+f@I&\\\"F1$\"+a\"yX]\"F1$\"+EHN0 :F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 122 "By looking at the sorted list, we see that the weights are between 8 and 16 kilograms. To get a visual sense of the data" }}{PARA 0 "" 0 "" {TEXT 226 16 "we can c reate a " }{TEXT 208 11 "histogram " }{TEXT 226 40 "of the data set. \+ We select a list of " }{TEXT 209 8 "ranges, " }{TEXT 226 173 " find the fraction of our 100 values that lie in each range, and then plot \+ a bar graph, where each bar stands over one of our ranges and the area of the bar is the fraction " }}{PARA 0 "" 0 "" {TEXT 226 356 "(no. of data values in the range/ 100). For example, we might start with th e ranges 8..10, 10..12, 12..14, 14..16. We will ignore the unlikely p ossibility of a data value falling exactly at an endpoint of one of th e ranges, so we specify the ranges by a single list: [8,10,12,14,16]. \+ We have defined a Maple procedure, histo, which does all the work: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "histo(data1,[8,10,12,14, 16]);" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 61 "We can get a more refined picture by making the ranges finer:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 40 "histo(data1,[8,9,10,11,12,13,14,15,16]);" }} {PARA 13 "" 1 "" {TEXT 228 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 273 "If we want to refine the ranges even more, it will be convenient to have Maple construct t he range list for us. Also, we will want to combine the histogram plo t with other plots later, so we give the histogram plot a name (termin ating with a colon : ) and then use the '" }{TEXT 210 7 "display" } {TEXT 226 67 "' command. We will do this for ranges of width 0.5 an d then 0.25" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "ranges1:= [s eq(8+.5*i,i=0..16)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(ranges1G73$ \"\")\"\"!$\"#&)!\"\"$\"#!*F+$\"#&*F+$\"$+\"F+$\"$0\"F+$\"$5\"F+$\"$: \"F+$\"$?\"F+$\"$D\"F+$\"$I\"F+$\"$N\"F+$\"$S\"F+$\"$X\"F+$\"$]\"F+$\" $b\"F+$\"$g\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "hist1dat a1:= histo(data1,ranges1):\n" }{MPLTEXT 1 0 24 " display(hist1data1 );" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "ranges2:= [seq(8+.25*i,i=0..32)];" }}{PARA 211 "" 1 " " {XPPMATH 20 "6#>%(ranges2G7C$\"\")\"\"!$\"$D)!\"#$\"$])F+$\"$v)F+$\" $+*F+$\"$D*F+$\"$]*F+$\"$v*F+$\"%+5F+$\"%D5F+$\"%]5F+$\"%v5F+$\"%+6F+$ \"%D6F+$\"%]6F+$\"%v6F+$\"%+7F+$\"%D7F+$\"%]7F+$\"%v7F+$\"%+8F+$\"%D8F +$\"%]8F+$\"%v8F+$\"%+9F+$\"%D9F+$\"%]9F+$\"%v9F+$\"%+:F+$\"%D:F+$\"%] :F+$\"%v:F+$\"%+;F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "hist 2data1:= histo(data1, ranges2): display(hist2data1);" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 226 374 "Notice that as we refine the ran ges the histogram becomes more jagged, even though the total area of t he bars is always 1. We would like to find a smooth looking graph whi ch represents these histograms in the sense that areas under the smoot h graph should be close to areas under the histogram graphs. For thi s purpose, mathematicians have invented various families of " }{TEXT 211 8 "density " }{TEXT 226 109 " functions, which form parametric f amilies. The most common and important of these is called the family \+ of " }{TEXT 212 20 "normal densities. " }{TEXT 226 70 "There are two parameters, m and s and the formula is given by normalp." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "normalp:=(x,m,s)->1/(sqrt(2*Pi)*s)* exp(-((x-m)/s)^2/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(normalpGf*6 %%\"xG%\"mG%\"sG6\"6$%)operatorG%&arrowGF**&-%$expG6#,$*(#\"\"\"\"\"#F 5*$),&F'F5F(!\"\"F6F5F5*$)F)F6F5F:F:F5*&-%%sqrtG6#,$*&F6F5%#PiGF5F5F5F )F5F:F*F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 285 "As you've seen in the text, these norma l densities have a bell shape. The parameter m gives the location of \+ the center of the bell and the parameter s measures the width of the \+ bell. The total area under the graph is 1, of course, and almost all \+ the area is between m-3s and m+3s. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 382 "Now we will try to find a normal density which smooths our histograms by plotting one of the histograms together with one of our normal densities. Mo re specifically, we want to find a normal density for which the areas \+ under the density over intervals on the line match the histogram areas . Thus we should pay more attention to the larger bars. We might sta rt with m=11, s=2. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "di splay([hist1data1,plot(normalp(x,11,2),x=8..16)]);\n" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 226 90 "It looks like we might be able to do better. Let's try m = 11.5, s=1 and then m=12, s=1.5." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "display([hist1data1,plot(normalp(x, 11.5,1),x=8..16)]);" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }}}{PARA 0 "" 0 "" {TEXT 226 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "display ([hist1data1,plot(normalp(x,12,1.5),x=8..16)]);" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }}}{PARA 0 "" 0 "" {TEXT 226 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 212 "" 0 "" {TEXT 200 132 " It turns out that the best values are about m=12 and s=1.5. Replace hi st1data1 in the command above with hist2data1 and compare. " }}{PARA 213 "" 0 "" {TEXT 213 6 "NOTE: " }{TEXT 200 14 " If we had " }{TEXT 214 14 "many thousands" }{TEXT 200 210 " of measurements instead of j ust 100, we would find that for any one of the histogram ranges we use d above, the normal density with m=12 and s=1.5 would match the histog ram very closely, as in the graph below. " }}{PARA 214 "" 0 "" {TEXT 200 0 "" }}{PARA 215 "" 0 "" {TEXT 200 595 " The biologist might now \+ use the density normalp(x,12,1.5) to describe the variation of the an imal's weight. For example, if she wanted to estimate the probabilit y that a randomly selected animal of this species will weigh between 9 and 13 kilograms, she might calculate the area under the density fun ction normalp(x,12,1.5) between 9 and 13. You might recognize that th e function giving this density does not have an elementary antiderivat ive so we had better resort to numerical integration. (Note: In actua l statistical practice we would estimate m and s arithmetically from t he data.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "evalf(Int(norm alp(x,12,1.5),x=9..13)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+0LdZs !#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 39 "Thus the probability is a bout .725 . " }}{PARA 0 "" 0 "" {TEXT 226 1 " " }}{PARA 0 "" 0 "" {TEXT 226 44 " It is often useful to use the " }{TEXT 215 41 "cumulative distribution function (c.d.f.)" }{TEXT 226 309 " i nstead of the density function to present information about the variab ility of our data. We find the probability of a value being between \+ two numbers, a and b, by integrating the density over the interval [a ,b]. The cumulative distribution function F(x) is defined as the int egral of the density from " }{XPPEDIT 18 0 "-infinity;" "6#,$%)infinit yG!\"\"" }{TEXT 226 57 " to x, so that F(x) gives the probability of v alue being " }{TEXT 216 14 "less than x. " }{TEXT 226 243 "For exampl e, the c.d.f. corresponding to the density normalp(x,12,1.5) is defin ed below, and its graph is plotted. (Remember also that the density f unction is the derivative of the corresponding c.d.f by the Fundamenta l Theorem of Calculus.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 " Fnormal:=(x,m,s)->int(normalp(t,m,s),t=-infinity..x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(FnormalGf*6%%\"xG%\"mG%\"sG6\"6$%)operatorG%&arr owGF*-%$intG6$-%(normalpG6%%\"tGF(F)/F4;,$%)infinityG!\"\"F'F*F*F*" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot( Fnormal(x,12,1.5), x= 7..16);" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 143 "If, for ex ample, we wanted the probability of a randomly chosen animal weighing less than 10 kilograms, we might estimate this probability by " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Fnormal(10,12,1.5);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"*)>7@\"*!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 4 "" 0 "" {TEXT 229 0 "" }} {PARA 4 "" 0 "" {TEXT 229 58 "Measuring central tendency: Means, Media ns and Percentiles" }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 132 " We often try to summarize distributions wi th a few numbers. Recall that the mean corresponding to a density p (x) is given by " }{XPPEDIT 18 0 "int(x*p(x), x = -infinity .. infinit y);" "6#-%$intG6$*&%\"xG\"\"\"-%\"pG6#F'F(/F';,$%)infinityG!\"\"F/" } {TEXT 226 94 ". For a normal density, the parameter m turns out to b e the mean as in our example above: " }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "int(x*normalp(x,12,1.5),x=-inf inity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+++++7!\")" }} }{EXCHG {PARA 0 "" 0 "" {TEXT 226 75 "(You can even do this integratio n by hand, using a substitution!) The " }{TEXT 217 17 "p-th perce ntile " }{TEXT 226 344 " is the number, which we will call xp for w hich the probability of a randomly chosen value being less than or eq ual to xp is p. Thus, xp is the solution of the equation F(xp) = p where p is a number between 0 and 1, expressed as a percentage, and F is the c.d.f. The 50-th percentile, x50, is what you already kno w as the median. " }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}{PARA 0 "" 0 " " {TEXT 226 262 " To find the 25-th, 50-th and 75-th percentiles , x25, x50 and x75, for the animal population discussed above, we so lve three equations using fsolve. We give fsolve a little help by sp ecifying a range of x values where it should look for a solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "x25=fsolve(Fnormal(x,12 ,1.5) = .25,x=7..16); x50=fsolve(Fnormal(x,12,1.5) = .50,x=7..16), x75 =fsolve(Fnormal(x,12,1.5)=.75,x=7..16);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%$x25G$\"+Ql#))4\"!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/%$x5 0G$\"+++++7!\")/%$x75G$\"+jM<,8F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 102 "Note that t he median and mean coincide for normal densities but not for many othe r kinds of densities." }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}{PARA 0 "" 0 "" {TEXT 226 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 218 10 "Problem 1. " }{TEXT 226 10 " The list " }{TEXT 219 6 "data2 " }{TEXT 226 86 " co ntains laboratory measurements of the lengths, in centimeters, of 80 e arthworms. " }}{PARA 0 "" 0 "" {TEXT 226 118 " a) Look at the data t o decide on appropriate histogram and plotting ranges and then plot a t least two histograms. " }}{PARA 0 "" 0 "" {TEXT 226 211 " b) Find \+ a normal density which gives good smoothing approximations to your his tograms. (Determine suitable values for m and s by trial and error.) \+ Plot your density on the same axes as one of your histograms." }} {PARA 0 "" 0 "" {TEXT 226 37 " c) Plot the corresponding c.d.f. " } }{PARA 0 "" 0 "" {TEXT 226 120 " d) Assuming that your density and c .d.f. accurately reflect the population of all earthworms, determine t he following" }}{PARA 0 "" 0 "" {TEXT 226 101 " i) The probabil ity that a randomly selected earthworm has length between 4 and 5 cent imeters." }}{PARA 0 "" 0 "" {TEXT 226 100 " ii) The probability \+ that a randomly selected earthworm has length more than 4.5 centimete rs." }}{PARA 0 "" 0 "" {TEXT 226 62 " iii) The 60th percentile, x 60, of earthworm lengths. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 0 " " }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}{PARA 0 "" 0 "" {TEXT 220 10 "Exa mple 2." }{TEXT 226 66 " Another important family of density functio ns is the family of " }{TEXT 221 11 "exponential" }{TEXT 226 60 " dens ities. These are specified by a single parameter, a, " }{TEXT 222 61 "and have value 0 when x is less than 0. For x greater than 0" } {TEXT 226 55 " the exponential density with parameter a is given by " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "exponp:=(x,a)-> a*exp(-a *x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'exponpGf*6$%\"xG%\"aG6\"6$% )operatorG%&arrowGF)*&F(\"\"\"-%$expG6#,$*&F'F.F(F.!\"\"F.F)F)F)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 226 129 "It's very important to remember \+ that this holds ONLY WHEN x is POSITIVE. Otherwise the density is 0. \+ Here's a plot, when a=0.5 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "plot(exponp(x,0.5),x =0..7);" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 28 "Since the density is 0 from " }{XPPEDIT 18 0 "-infinity; " "6#,$%)infinityG!\"\"" }{TEXT 226 81 " to 0, the c.d.f. Fexpon(x) c an be found by integrating from 0 to x instead of " }{XPPEDIT 18 0 "- infinity;" "6#,$%)infinityG!\"\"" }{TEXT 226 127 " to x. You should be able to do this by hand, but we'll have Maple do it here, and then we 'll plot the c.d.f. when for a=0.5. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Fexpon:=(x,a)-> int(exponp(t,a), t=0..x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'FexponGf*6$%\"xG%\"aG6\"6$%)operatorG%&arro wGF)-%$intG6$-%'exponpG6$%\"tGF(/F3;\"\"!F'F)F)F)" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 47 "value(Fexpon(x,a));plot(Fexpon(x,0.5), x=0.. 7);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%$expG6#,$*&%\"xG\"\"\"%\"aG F*!\"\"F,F*F*" }}{PARA 13 "" 1 "" {TEXT 228 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 11 " The list " }{TEXT 223 5 "data3" }{TEXT 226 270 ", which we display below, consists of 80 measurements of the time elapsed, in minutes, betwee n the arrival of phone calls at a certain 800 phone number. There ar e theoretical reasons to assume that the distribution of such times is given by an exponential density. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "data3;" }}{PARA 216 "" 1 "" {XPPMATH 20 "6#7go$\"+$*)p .7\"!#5$\"+\"yRTL#!\"*$\"+Ir;yYF)$\"+%>s`q%F&$\"+w3yGBF)$\"+\"*f#[B\"! \")$\"+n)4U,*F)$\"+;q=OaF)$\"+(Q[P\"**F)$\"+2c;S:F)$\"+Zi;w>F2$\"+g^<, ZF)$\"+?%fx!**F&$\"+Xsou@F)$\"+S`+y;F2$\"+-%=Y`%F2$\"+Ss)e`\"F)$\"+Msv C[F&$\"+nr)3V\"F2$\"+'y![jkF)$\"+@5MSF2 $\"+Zw9J$)F&$\"+I9&3r$F)$\"+BbpniF&$\"+(>:#=lF&$\"+u!y*R(*!#6$\"+#e$Qw XF)$\"+r2zh@F2$\"+[,u/XF)$\"+TSI1HF2$\"+$4;.-#F)$\"+tc(=j$F)$\"+kaB()> F2$\"+ltrN`F&$\"+Dn3m:F)$\"+Ap2j>F)$\"+!))[&fpF)$\"+nGp!\\#F)$\"+b\\B2 HF&$\"+aM)pJ)F)$\"+CXb!o\"F2$\"+LRX&F)$\"+HV3O9F&$\"+>b0ODF2$\"+\"G \"\\'o$F)$\"+3+ahRF)$\"+X&e2Y\"F2$\"+C!Qmh\"F2$\"+a'>M7\"F2$\"+)GmLc#F )$\"+rz!o)GF&$\"+ksmiVF)$\"+8,s)R$F)$\"+aT2-(*F)$\"+gK%>O'F)$\"+XB/W \\F)$\"+>j()>EF2$\"+)=!p#z&F&$\"+53&31\"F2$\"+MK+J$)F)$\"+[ls1@F2" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 224 9 "Problem 2" }{TEXT 226 232 ". a) Plot some histograms, s tarting with ranges [0,2,4,...,30] and find an exponential density tha t appears to smooth the histograms. Give your value for the paramete r a. (Hint: Start with a=.5. Plot over the interval [0,20].) " }} {PARA 0 "" 0 "" {TEXT 226 133 " b) Assuming that t he density you found in a) is reflective of the typical times between \+ phone calls, answer the " }}{PARA 0 "" 0 "" {TEXT 226 21 "following qu estions. " }}{PARA 0 "" 0 "" {TEXT 226 133 " \+ i) What is the probability that at least 8 minutes elapse between the most recent phone call and the next?" }}{PARA 0 "" 0 "" {TEXT 226 134 " ii) What is the probability that the ti me between two successive phone calls is between 2.5 and 6.5 minutes?" }}{PARA 0 "" 0 "" {TEXT 226 65 " iii) What is \+ the mean time between calls?" }}{PARA 0 "" 0 "" {TEXT 226 67 " \+ iv) What is the median time between calls?" }}{PARA 0 "" 0 "" {TEXT 226 91 " v) What is the 75th p ercentile of the elapsed time between calls?" }}{PARA 0 "" 0 "" {TEXT 226 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 225 10 "Problem 3." }{TEXT 226 52 " Consider the function, DEFINED WHEN x > 0, by " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "pnotadensity:=x->x^(5/2)*exp(-x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-pnotadensityGf*6#%\"xG6\"6$%)operatorG%&arrowGF(*&) F'#\"\"&\"\"#\"\"\"-%$expG6#,$F'!\"\"F1F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 104 "and de fined to be 0 when x < 0. a) Determine a value for k so that the fu nction DEFINED WHEN x > 0 by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "p:=x->k*pnotadensity(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\" pGf*6#%\"xG6\"6$%)operatorG%&arrowGF(*&%\"kG\"\"\"-%-pnotadensityGF&F. F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 226 195 "and defined to be 0 w hen x < 0, IS a density. (Remember that densities must have integral 1 .) Put the value you find for k into the definition of p and use it \+ for the remainder of this problem." }}{PARA 0 "" 0 "" {TEXT 226 140 " \+ b) Plot the density function p(x) and the \+ corresponding c.d.f., for x>0 of course, on the same axes if you can." }}{PARA 0 "" 0 "" {TEXT 226 84 " c) Use M aple to find the corresponding mean and median." }}{PARA 0 "" 0 "" {TEXT 226 129 " d) If a random value has t his density, find the probability that the value lies between 1.5 and \+ 3.2." }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}}{EXCHG {PARA 4 "" 0 "" {TEXT 229 17 "Homework Problems" }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}{PARA 0 "" 0 "" {TEXT 226 69 " Your homework for this worksheet consists of Problems 1,2,3 above." }}{PARA 0 "" 0 "" {TEXT 226 0 "" }}}{EXCHG {PARA 217 "" 0 "" {TEXT 200 77 "MTH 142 Maple Worksheets written by B. Kaskosz and L. Pakula, Copyright 1999." }}}{EXCHG {PARA 218 "" 0 "" {TEXT 200 26 "Last modified August 1999." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }