1. Mean = 2025.25

Median = 2209

Mode = 0

  1. Mean = 41.4

Median = 40.76

Mode = 0

  1. Mean = 3670

Median = 3830

Mode = 4090

  1. Mean = 266

Median = 266

Mode = 260

    1. Mean > Median

  1. Mean = Median

  2. Mean < Median

  1. Sample Variance = 36; Standard Deviation = 6

  2. Sample Variance = 100; Standard Deviation = 10

  3. Population Variance = 16; Standard Deviation 4

  4. Population Variance = 64; Standard Deviation = 8

  5. Sample Variance = 196; Standard Deviation = 14

  6. Population Variance = 25; Standard Deviation = 5

  7. Range = 25 Sample Standard Deviation = 8.43

    1. Sample Standard Deviation = 6; Mean = 104
  1. Yes, we can use this rule because the histogram seems bell shaped.
  2. 95%
  3. 93%
  4. 16%
  5. 12%

  1. The Z score for the 32-35 weeks baby is -0.3 and that for the 40 weeks is -0.43. The 40 weeks baby weighs less.

  2. The 75 inch man

  3. Johnson, because the Z value was greater

  4. Dario’s car had the most convining victory

  5. 239

15)a) 15% of males between 3 to 5 months of age have a head circumference lower than 41 cm.

  1. 90% of the females of 2 years of age have a waist circumeference lower than 52.7 cm

  2. 10% or lower of age group 20-29 have height as 166.8, 25% or lower have 171.5, 50% or lower have 176.7, 75% or lower have 181.4 and 90% or lower have 186.8 - and so on.

    1. The z score is -1.21. This means that it is 1.21 standard deviations from the mean, towards the left
  1. Q2 = 9.95 ; Q1 = 9.15 ; Q3 = 11.1
  2. IQR = 1.95
  3. UP = 14.025 ; LF = 6.225. Yes. 14.025 is an outlier.

  1. Q2 = 466; Q1 = 433 ; Q3 = 489.5 IQR = 56.5 UF = 574.25

  2. Right-skewed distribution Median/Q2 = 3 ; Q1 = 1 ; Q3 = 6 Min = 0 ; Max = 16

4)Symmetric Distribution Median/Q2 = 5; Q1 = 2 ; Q3 = 7 Min = -1 ; Max = 11

    1. 40
  1. 52
  2. y because the IQR is greater
  3. Symmetric Distribution - The length from the median to min and the length from the median to max are similar.
  4. Right-skewed distribution because the length from the median to max is greater than the length from the median to the min.
    1. 16
  1. 22
  2. y will have more dispersion because of a greater IQR
  3. yes, it does and the value is 30
  4. Left skewed distribution because the length from the median to the min is greater than the length from the median to the max
my_data <- c(60, 68, 77, 89, 98)

boxplot(my_data, horizontal = T)

fivenum(my_data)
## [1] 60 68 77 89 98
max(my_data)
## [1] 98
min(my_data)
## [1] 60
my_data <- c(110, 140, 157, 173, 205)

boxplot(my_data, horizontal = T)

fivenum(my_data)
## [1] 110 140 157 173 205
max(my_data)
## [1] 205
min(my_data)
## [1] 110
  1. Q2 = 54.5 ; Q1 = 51 ; Q3 = 57.5 Min = 42 ; Max = 69
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)

boxplot(my_data, horizontal = T)

fivenum(my_data)
## [1] 42.0 50.5 54.5 57.5 69.0
max(my_data)
## [1] 69
min(my_data)
## [1] 42
  1. Symmetric Distribution with an outlier
    1. Q2 = 10; Q1 = 9.05; Q3 = 11.2 Max = 16.4; Min = 7.2
my_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)

boxplot(my_data, horizontal = T)

fivenum(my_data)
## [1]  7.20  9.05 10.00 11.20 16.40
max(my_data)
## [1] 16.4
min(my_data)
## [1] 7.2
  1. Right skewed with one outlier

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