3.4
5.-0.30; -0.43; the 40 week’s gestation baby weights less relative to the gestation period.
The man is relatively taller because it got bigger z socres.
Clayton Kershaw had the better year because his ERA was 2.30 standard deviations below the National League mean ERA, but Hernandez???s ERA was 1.91 standard deviations below the American League???s mean ERA. The Clayton Kershaw is more standard deviations below the mean so it performs better.
Ryan is better at the 100 meters’race. Because in 100 meters’race, Ryan’s z score is 3.39 below the standard deviation blow while in 200 meters’race, his z score is 3.05 standard deviation below. He performed better in the 100 meters’race.
239
15
a). 15% of 3- to 5-month-old males have a head circumference that equals to 41.0 cm or less, and 85% of 3- to 5-month-old males have a head circumference that is bigger than 41.0 cm.
c). The heights at each percentile decrease as the age increases. This menas that adult males are getting taller than the middel and older ages adults.
22
a). Based on the calculation, the result will be 1.24 standard deviation below the average.
b). Q1=9.15 Q2=9.95 Q3=11.1
c). Q3-Q1=1.95
d)UF=14.025 Lf=6.225. The upper fence is a outlier while the lower fence is not a outlier.
3.5
3
a)Skewed Right
b)Minimum:0 Q1=1 Q2=3 Q3=6 Maximum:16
4
a)Symmetric
b)Minimum:-1 Q1=2 Q2=5 Q3=8 Maximum:11
5
a).40
b).52
c).y. Because y is more wider in the graph and contains more values.
d)Symmetric. Because we can see in the graph, the shape of x is a almost bell-curved shape. The Q2 value lay in the middle of the graph.
e)Skewed Right. Because the biggest value is in the left side and more values are dispersed into right side.
6
a).16
b).22
c).y. Because y is more wider in the value dividation.
d)yes. The first one is 37 and the other one is -3.
e)Skewed left. Because the biggest value is in the right and the more values are divided in the left side. It is a Skewed left shape.
7
dat1 <- c(60,68,77,89,98)
boxplot(dat1)
8
dat2 <- c(110,140,157,173,205)
boxplot(dat2)
9
dat3 <- c(42,43,46,46,47,
47,48,49,49,50,
50,51,51,51,51,
52,52,54,54,54,
54,54,55,55,55,
55,56,56,56,57,
57,57,57,58,60,
61,61,61,62,64,
64,65,68,69)
a)42,51,55,58,69
boxplot(dat3)
c). symmetric. Because as the shape shows, the median is almost in the middle of the boxplot and values are evenly sivided.so it will be a symmetric shape