Math 203 Assignment 1.pdf - MATH 203 Assignment 1 Problem...

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MATH 203 Assignment 1Problem2.31,a)It is a Histogram because the horizontal axis is divided into several intervals withsame width and vertical bars are placed over each body length class interval todisplay the frequency or different body length class relative frequencies shown asheight of the vertical of the bar.b)For height between 87 and 91 centimeters, there are 16+22=38 armadillos, byadding heights of bars with label 87-89, 89-91.c)In total, 80 armadillos were captured and there are 38 armadillos with body lengthbetween 87 and 91 centimeters, so the proportion is 38 out of 80 (38/80=0.475).d)For heights of armadillos between 83 and 93 centimeters, there are8+16+22+10+13 =69 armadillos legal to be captured. Because total of heights ofbars between dotted lines=69, it means that 80-69=11 out of 80 capturedarmadillos are illegal (11/80ā‰ˆ0.14).e)There are 16 armadillos in the interval of 87-89cm and 34 armadillos more than orequal to 89cm. Even though the exact number of armadillos more than 88cm couldnot be determined by the histogram, the range would be [34, 50] for number ofarmadillos more than 88cm. In another word, the proportion would be between34/80=0.425 and 50/80=0.625 because the highest value would be achieved if 16armadillos are all more than 88cm in interval of 87-89cm or there would be noarmadillos in that interval leading to the least value of armadillos (34 out of 80).f)The type of skewness present in these data is skewed to the left not only due tothe left tail is longer and the mass of the distribution is concentrated on the rightof the figure but also because the median (the 40thand 41stterms) of 80 armadilloslies between 87-89, approximately M=12(87 +1016Ɨ 2 + 87 +1116Ɨ 2ļ¼‰=88.31while the mean, using histogram mean estimation,š‘„Ģ… =š‘„1+š‘„2+ā‹Æ+š‘„š‘›?=āˆ‘š‘„š‘–š‘›š‘–=1?=(78)(2)+(80)(4)+(82)(3)+(84)(8)+(86)(13)+(88)(16)+(90)(22)+(92)(10)+94+9680= 87.63, so themean is smaller than the median which, in other word, indicates skewed to the left.
Problem2.62a)Since there are 35 improvement scores for the honey dosage group, the medianformula for sample size of odd number is M=š‘„š‘›+12āˆ—. Using Excel to calculate themedian improvement score for the honey dosage group, M=11.b)Since there are 33 improvement scores for the DM dosage group, the medianformula for sample size of odd number is M=š‘„š‘›+12āˆ—. Using Excel to calculate themedian improvement score for the DM dosage group, M=9.,š‘›

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Term
Summer
Professor
Dr.JoseCorrea

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