Homework #2 -- Suggested Answers

ISBE 2.6, 2.10-2.12, 2.19, 2.23, 2.29

 

 

2.6

a. NOTE: These plots were produced using Microsoft Excel and are meant to give you an idea of what your graph should look like, you should have drawn your versions by hand.

2.6 a.

 

b. Median = 11

The first quartile is 8 and the third is 39

c.

 

 

2.10

a.

b. Mean = 70.75, Mode = 75

 

2.11

a.

b. Mean = 71.5, Mode = 80

 

 

2.12

Grouping

Mean

Mode

Original Data

70.4

Not Defined

Fine Grouping

70.75

75

Coarse Grouping

71.5

80

 

  1. The mode is not a very good measure of central tendency since it depends on the arbitrary grouping of the data.
  2. The fine grouping gives a closer approximation to the mean of the original data.

 

2.19

a. A 3-D plot like this is nice, but not necessary. 3 separate histograms, stacked one above the other, or 3 box plots next to each other would do the trick.

 

Yes, there appears to be a downward trend. Lower viscosities are more frequent by day 3. In addition, the data become less variable as time passes.

 

b.

 

Yes, both the mean and standard deviation decrease as time passes, the former reflecting the downward trend in viscosities, the later reflecting the decrease in variability in viscosity measurements.

 

c.

Mean

Standard Deviation

Day 1

74.8

4.84

Day 2

71.3

4.14

Day 3

70.3

2.75

 

The mean for the 150 observation is 72.13 and the standard deviation is 4.43. The mean is related to the above in that it equals the average of the three day means. The overall standard deviation is larger than the average of the three daily standard deviations.

 

2.23

  1. The average length is 40 feet.
  2. The total area is 12,000 square feet and the average area is 2,400 square feet.
  3. No (the square of a sum of numbers divided by their number does not, in general, equal the sum of squares of those numbers divided by their number).

 

2.29

  1. The Berezina River -- the army went from 50,000 men to 28,000. It was about -20 degrees (centigrade).
  2. Two groups split off early: 30,000 were added at Bota and 6,000 added near the end.
  3. A soldier had about a 24% chance to make it Moscow alive. A soldier returning to Poland from Moscow had a (20,000/100,000)= 20% chance of making it as far as Bota. Those that made it that far had a (4,000/50,000) = 8% chance of returning. The chance of returning from Moscow is 0.20*0.08 = 1.6% (the book seems to be in error, they estimate 2.5%). Remember that of the 10,000 soldiers that managed to return to Poland a portion of them were among the 36,000 soldiers from question b.

 

2.

  1. The five number summary: median = 3.8, first quartile = 3.575, third quartile = 4.4, the lowest individual value = 2.2, the highest individual value = 9.6, IQR = 0.825
  2. The mean = 4.191667, the standard deviation = 1.81932
  1. The mean = 3.7, the standard deviation = 0.67082
  2. The five number summary: median = 3.7, first quartile = 3.55, third quartile = 4.15, the lowest individual value = 2.2, the highest individual value = 4.5, IQR = 0.6
  3. Percentile-based measures (median and IQR) are more robust statistics--they are more resistant to a change from a wild swing in a single observation.