3.4 Homework

  1. 34 weeks z= -0.30 40 weeks z= -0.43 The 40 week old baby weights less relative to the gestation period because the number of standard deviations is further from the mean gestation period (more negative)

  2. The man is relatively taller with a z score of 1.8 because he is more standard devaiations above the mean height for men when compared to the woman. She is relatively less standard deviations above the mean relative to her mean with a z score of 1.55

  3. Hernandez from the American League performed better in his league than the other player from the National league relative to his mean. Hernandez had a z score of -2.14 which is less than the mean earned run average for that league. Being more standard deviations below the mean is better because it means he performed that many standard deviations better than the mean. Relatively speaking, the other player only had a z score of -1.78 which is relatively closer to his mean in his league.

  4. Power had the more convincing victory in the Sonoma race because his z score is -1.756 which is comparatively more standard deviations away from the mean for that race than the other driver finished with in his race. The other driver, Franchitti only finished -1.47 standard deviations below his mean finish time for the Indy 500 race. Power did better because his time was futher below the mean, so it was a faster time relative to the mean.

  5. 1.5 = x - 200/26 x = 239 The minimus score that an applicant must get on the test is a 239.

  1. means that 15% of observations recorded for males 3-5 months old had a head circumferance smaller than/equal to 41.0cm while 75% had a larger head circumference

  2. means that 90% of observations for waist circumference of females at 2 years old had a circumference of 52.7 or less while 10% had a larger waist circumference

  3. Observation of how body meansurements of men change overtime - height decreases with age regardless of the percentile

  1. z score of Blackie relative to average hemoglobin = -1.21. This means that Blackie the cat has a hemoglobin reading 1.21 standard deviations less than the mean hemogolbin of all cats sampled

  2. Q1= 9.55 Q2= 9.95 Q3= 11.1

  3. IQR = Q3-Q1 = 1.55

  4. LF= 9.55 - 1.5(1.55) = 7.23 UF= 11.1 + 1.5(1.55) = 13.43 There is one outlier that is lower than the lower fence *5.7 g/dL

  1. Q1= 433 Q2= 466 Q3= 489.5 IQR=56.5

    UF= 489.5 + 1.5(56.5) UF= 574.25 minutes should be the cut off and customers should be contacted

3.5 Homework

  1. Min=0 Q1=1 Q2=3 Q3=6 Max=16 The data is right skewed because the space between Q3 and the max acts as the tail in a graph, the vaules are increasing but the frequency is less, as with a histogram.

  2. min= -1 Q1=2 Q2=5 Q3=8 max=12 The data here is symmetric the median (Q2) is an equal distance from the min and the max.

  1. Median of variable x = 40

  2. Q3 of variable y = 53

  3. variable y has more dispersion because the interquartile range (IQR) is greater than variable x

  4. variable x is symmetric because it has a Q1, Q2, and Q3 that are spread equally with the whiskers equal as well.

  5. variable y is right skewed because the whisker on the right side indicates a lot of space bewteen Q3 and the max which represents the descending tail of a right skewed graph

  1. median of variable x = 16

  2. Q1 of variable y = 22

  3. Variable y has more dispersion because its IQR is greater than variable x

  4. variable x has 30 as an outlier

  5. variable y is a left skewed shape because the distance between Q2 and Q1 and Q1 to the min is greater indicating it has more observations on that side to be recorded making the graph shape downward like that.

your_data <- c(60, 63, 68, 68, 68,
               75, 75, 77, 79, 89, 
               89, 89, 93, 94, 98)
             
boxplot(your_data, horizontal = T)

fivenum(your_data)
## [1] 60 68 77 89 98
this_data <- c(110, 125, 140, 140,140,
               150, 152, 157, 160, 173, 
               173, 173, 180, 180,205)
             
boxplot(this_data, horizontal = T)

fivenum(this_data)
## [1] 110 140 157 173 205

9

my_data <- 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)

fivenum(my_data)
## [1] 42.0 50.5 54.5 57.5 69.0
boxplot(my_data, horizontal = T)

9c. The shape of this box plot appears to be almost symmetrical. The space between Q1 and Q2 is a little larger than bewteen Q2 and Q3 but there is no dramatic shift on either side of the graph indicating it is roughly symmetrical (data does contain an outlier though)

our_data <- c(7.2, 7.8, 7.8, 7.9, 8.1, 8.3,
              8.5, 8.6, 8.6, 8.6, 8.7, 8.8,
              9.0, 9.1, 9.2, 9.2, 9.2, 9.4,
              9.4, 9.6, 9.7, 9.7, 9.9, 9.9,
              10.0, 10.0, 10.0, 10.1, 10.2,10.3,
              10.3, 10.3, 10.3, 10.7, 10.7, 10.9,
              11.2, 11.2, 11.2, 11.3, 11.3, 11.3,
              11.5, 11.5, 11.7, 12.4, 12.5, 13.6,
              13.8, 14.4, 16.4)


fivenum(our_data)
## [1]  7.20  9.05 10.00 11.20 16.40
boxplot(our_data, horizontal = T)

10c. The shape of this box plot appears to be right skewed. The larger space between Q2 and Q3 and the space from Q3 to the max indicates more data observations on that side of the graph which means a tail descending to the right side of the graph. (also contains an outlier)