Maiko YoshidaGOLD MEDAL HEIGHTSThe table below gives the height (in centimeters) achived by the gold medalists at variousOlympic Games.Year 1932 1936 1948 1952 1956 1960 1964 1968 1972 1976 1980Height 197 203 198 204 212 216 218 224 223 225 236(cm)Using technology, plot the data points on a graph. Define all variables used and stateany parameters clearly. Discuss any possible constraints of the task.Note: The Olympic Games were not held in 1940 and 1944.Year 8 9 12 13 14 15 16 17 18 19 20Height 1.97 2.03 1.98 2.04 2.12 2.16 2.18 2.24 2.23 2.25 2.36(cm)X=1900+4xY=100y
Maiko YoshidaIf the numbers were big, a computational error would be big, as well. In order to avoidthis, and also to calculate easily, I scaled the numbers.Year: X=1900+4x Ex. 1932=1900+4(8) x=8Height: Y=100y Ex. 197=100(1.97) y=1.97Parametersy ax 2 bx ca represents a direction and how the graph opens For example, if a is positive value, it means the quadratic graph open to upward. bthe relationship between a and b is x 2ac represents y-intersectPossible constraints are that X can only be 8,9,10,11,12,13,14,15,16,17,18,19,20.y can only be positive real number.What type of function models the behaviour of the graph? Explain why you chosethis function. Analytically create an equation to model the data in the above table.The height once decreases but basically it increases, as years pass so I think quadraticequation is the best fit. I picked three examples up to make a manual equation to modelthe data. I used 1932, 1960 and 1980 because these are at the first year, last year andmiddle year of the data so it can be said that these three points are typical of this graph.For creating the equation, I used matrix.First year (1932) 1.97=64a+8b+cMiddle year (1960) 2.16=225a+15b+cLast year (1980) 2.36=400a+20b+c
Maiko Yoshida 64 8 1 a 1.97225 15 1 b = 2.16400 20 1 c 2.38y 0.0010714286 x 2 0.0025 x 1.881428571On a new set of axes, draw your model function and the original graph. Commenton any differences. Discuss the limitations of your model. Refine your model ifnecessary.My model function almost fits the original dots but some point such as x=12 or 13 do notsuit to the graph. In order to refine the graph, I focused on the turning point. In this graph,there are 4 turning points. Therefore, I made quintic equation.y ax 5 bx 4 cx 3 dx 2 ex f
Maiko YoshidaThis equation seems much better to fit the graph within x=8 to 20.Use technology to find another function that models the data. On a new set of axes,draw both your model functions. Comment on any differences.I tried several equations by using logger-pro and checked the RMSE, which means RootMean Square Error. The smaller this value, the graph is much better fit.y ax 2 bx c Quadratic function RMSE=0.03868
Maiko Yoshiday= a bx cx 2 dx 3 Cubic function RMSE=0.03942y aln(bx) Natural log function RMSE=0.05681
Maiko Yoshiday alog(bx) Base-10 logarithm RMSE=0.05681y mx b Linear function RMSE=0.04523Because RMSE of quadratic function is the smallest value, it can be said that quadraticequation is the best fitted function.
Maiko YoshidaThe graph which shown above is the auto fit quadratic functionThe graph which shown above is the manual quadratic function.
Maiko YoshidaThe graph which two graphs combined is shown below.The RMSE of manual equation is 0.0480387.The RMSE of auto fit equation is 0.03868.Although there are some errors between manual equation and auto fit equation, it seemsalmost same because two of RMSE are nearly same, as well.
Maiko YoshidaHad the Games been held in 1940 and 1944, estimate what the winning heightswould have been and justify your answers.Year: 1940If I use the manual fit quadratic equation, the formula would be1940 1900 4 xx 10y 0.00107 102 0.0025 10 1.881y 2.013Y 100 2.013 201.3 A. 201.3 cmIf I use the auto fit quadratic equation, the formula would be1940 1900 4 xx 10y 0.00181 102 0.0203 10 2.027y 2.005Y 100 2.005 200.5 A. 200.5cmYear: 1944If I use the manual fit quadratic equation, the formula would be1944 1900 4 xx 11y 0.00107 112 0.0025 10 1.881y 2.035Y 100 2.035 203.5 A. 203.5cm
Maiko YoshidaIf I use the auto fit quadratic equation, the formula would be1944 1900 4 xx 11y 0.00181 112 0.0203 11 2.027y 2.02Y 100 2.022 202.2 A. 202.2cmThe answers are appropriate because the answers which I got are almost same value as thegraph above.Use your model to predict the winning height in 1984 and in 2016. Comment onyour answers.Year: 1984If I use the manual fit quadratic equation, the formula would be1984 1900 4 xx 21y 0.00107 212 0.0025 21 1.881y 2.405Y 100 2.405 240.5 A. 240.5cmIf I use the auto fit quadratic equation, the formula would be1984 1900 4 xx 21y 0.00181 212 0.0203 21 2.027y 2.398Y 100 2.398 239.8 A. 239.8cm
Maiko Yoshida Year: 2016 If I use the manual fit quadratic equation, the formula would be 2016 1900 4 x x 29 y 0.00107 292 0.0025 29 1.881 y 2.853 Y 100 2.853 285.3 A. 285.3cm If I use the auto fit quadratic equation, the formula would be 2016 1900 4 x x 29 y 0.00181 292 0.0203 29 2.027 y 2.961 Y 100 2.961 296.1 A. 296.1cm As year passes, the value of height is going to be unbelievable number because the human will never achieve 285 or 296.1cm. Therefore, the quadratic function gradually cannot be used as the year passes. The following table gives the winning heights for all the other Olympic Games since 1986.Year 1896 1904 1908 1912 1920 1928 1984 1988 1992 1996 2000 2004 2008Height 190 180 191 193 193 194 235 238 234 239 235 236 236(cm) I scaled again.
Maiko YoshidaYear -1 1 2 3 5 7 21 22 23 24 25 26 27Height 1.90 1.80 1.91 1.93 1.93 1.94 2.35 2.38 2.34 2.39 2.35 2.36 2.36(cm)How well does your model fit the additional data?As graph shows, the quadratic equation does not fit the dots very well. If I use thequadratic equation, human can achieve 280cm in 2060. Obviously, human cannot jump280cm so the equation which I used has restriction. Therefore, it can be said that this isinterpolated example.
Maiko YoshidaDiscuss the overall trend from 1896 to 2008, with specific references to significantfluctuations.At first, the slope of the graph seems constant except in 1904. However, from 1936 to1980, the sloop becomes steep. Then, the slope seems to be constant again from 1980.
Maiko YoshidaWhat modifications, if any, need to be made to your model to fit the new data?The graph looks like arc tangent which function is constant at first, then becomes sharpand be constant again. Because the ability of human’s jump does not increase forever, itshould be stable in the end. Therefore, I will use a atan x b even though the value ofRMSE is big.