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

MATH 203 Assignment 1
Problem 2.31 ,
a) It is a Histogram because the horizontal axis is divided into several intervals with
same width and vertical bars are placed over each body length class interval to
display the frequency or different body length class relative frequencies shown as
height of the vertical of the bar.
b) For height between 87 and 91 centimeters, there are 16+22=38 armadillos, by
adding heights of bars with label 87-89, 89-91.
c) In total, 80 armadillos were captured and there are 38 armadillos with body length
between 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 are
8+16+22+10+13 =69 armadillos legal to be captured. Because total of heights of
bars between dotted lines=69, it means that 80-69=11 out of 80 captured
armadillos are illegal (11/80≈0.14).
e) There are 16 armadillos in the interval of 87-89cm and 34 armadillos more than or
equal to 89cm. Even though the exact number of armadillos more than 88cm could
not be determined by the histogram, the range would be [34, 50] for number of
armadillos more than 88cm. In another word, the proportion would be between
34/80=0.425 and 50/80=0.625 because the highest value would be achieved if 16
armadillos are all more than 88cm in interval of 87-89cm or there would be no
armadillos 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 to
the left tail is longer and the mass of the distribution is concentrated on the right
of the figure but also because the median (the 40 th and 41 st terms) of 80 armadillos
lies between 87-89, approximately M =
1
2
(87 +
10
16
× 2 + 87 +
11
16
× 2 ） = 88.31
while the mean, using histogram mean estimation, 𝑥̅ =
𝑥
1
+𝑥
2
+⋯+𝑥
𝑛
?
=
∑
𝑥
𝑖
𝑛
𝑖=1
?
=
(78)(2)+(80)(4)+(82)(3)+(84)(8)+(86)(13)+(88)(16)+(90)(22)+(92)(10)+94+96
80
= 87.63 , so the
mean is smaller than the median which, in other word, indicates skewed to the left.

Problem 2.62
a) Since there are 35 improvement scores for the honey dosage group, the median
formula for sample size of odd number is M= 𝑥
𝑛+1
2
∗
. Using Excel to calculate the
median improvement score for the honey dosage group, M=11.
b) Since there are 33 improvement scores for the DM dosage group, the median
formula for sample size of odd number is M= 𝑥
𝑛+1
2
∗
. Using Excel to calculate the
median improvement score for the DM dosage group, M=9.
,
𝑛