Assignment # 1 Solution – Statistics

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Assignment # 1 Solution - Statistics

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Assignment # 1 Solution
1.1 a The experimental unit, the individual or object on which a variable is measured, is the student.
b The experimental unit on which the number of errors is measured is the exam.
c The experimental unit is the patient.
d The experimental unit is the azalea plant.
e The experimental unit is the car.
1.2 a “Time to assemble” is a quantitative variable because a numerical quantity (1 hour, 1.5 hours, etc.) is
measured.
b “Number of students” is a quantitative variable because a numerical quantity (1, 2, etc.) is
measured.
c “Rating of a politician” is a qualitative variable since a quality (excellent, good, fair, poor) is
measured.
d “Province or territory of residence” is a qualitative variable since a quality (ON, AB, BC, etc. ) is
measured.
1.3 a “Population” is a discrete variable because it can take on only integer values.
b “Weight” is a continuous variable, taking on any values associated with an interval on the real line.
c “Time” is a continuous variable.
d “Number of consumers” is integer-valued and hence discrete.
1.9 a The variable “reading score” is a quantitative variable, which is usually integer-valued and hence
discrete.
b The individual on which the variable is measured is the student.
c The population is hypothetical—it does not exist in fact—but consists of the reading scores for all
students who could possibly be taught by this method.
1.17 The most obvious choice of a stem is to use the ones digit. The portion of the observation to the right of the
ones digit constitutes the leaf. Observations are classified by row according to stem and also within each
stem according to relative magnitude. The stem and leaf plot is shown below.
1 | 6 8
2 | 1 2 5 5 5 7 8 8 9 9
3 | 1 1 4 5 5 6 6 6 7 7 7 7 8 9 9 9 leaf digit = 0.1
4 | 0 0 0 1 2 2 3 4 5 6 7 8 9 9 9 1 2 represents 1.2
5 | 1 1 6 6 7
6 | 1 2
a The stem and leaf plot has a mound-shaped distribution.
b From the stem and leaf plot, the smallest observation is 1.6 (1 6).
c The eight and ninth largest observations are both 4.9 (4 9).
1.18 a For n = 50, use between 8 and 10 classes.
b
Class i Class Boundaries Tally fi Relative frequency, fi/n
1 1.6 to < 2.1 11 2 0.04
2 2.1 to < 2.6 11111 5 0.10
3 2.6 to < 3.1 11111 5 0.10
4 3.1 to < 3.6 11111 5 0.10
5 3.6 to < 4.1 11111 11111 1111 14 0.28
6 4.1 to < 4.6 11111 1 6 0.12
7 4.6 to < 5.1 11111 1 6 0.12
8 5.1 to < 5.6 11 2 0.04
9 5.6 to < 6.1 111 3 0.06
10 6.1 to < 6.6 11 2 0.04
Relative Frequency
1.6 2.1 2.6 3.1 3.6 4.1 4.6 5.1 5.6 6.1 6.6
0.30
0.20
0.10
0
c From b, the fraction less than 5.1 is that fraction lying in classes 1–7, or
(2 + 5 +L+ 7 + 5) 50 = 43 50 = 0.86
d From b, the fraction larger than 3.6 lies in classes 5–10, or
(14 + 7 +L+ 3 + 2) 50 = 33 50 = 0.66
e The stem and leaf plot has a more peaked mound-shaped distribution than the relative frequency
histogram because of the smaller number of groups.
1.25 a The range of the data 32.3-0.2 = 32.1. We choose to use eleven class intervals of length 3
(32.1/11 = 2.9, which when rounded to the next largest integer is 3). The subintervals 0.1 to < 3.1,
3.1 to < 6.1, 6.1 to < 9.1, and so on, are convenient and the tally is shown below.
Class i Class Boundaries Tally fi Relative Frequency, fi/n
1 0.1 to < 3.1 11111 11111 11111 15 15/50
2 3.1 to < 6.1 11111 1111 9 9/50
3 6.1 to < 9.1 11111 11111 10 10/50
4 9.1 to < 12.1 111 3 3/50
5 12.1 to < 15.1 1111 4 4/50
6 15.1 to < 18.1 111 3 3/50
7 18.1 to < 21.1 11 2 2/50
8 21.1 to < 24.1 11 2 2/50
9 24.1 to < 37.1 1 1 1/50
10 27.1 to < 30.1 0 0/50
11 30.1 to < 33.1 1 1 1/50
The relative frequency histogram is shown below.
Relative frequency
0 10 20 30
15/50
10/50
5/50
0
b The data is skewed to the right, with a few unusually large measurements.
c Looking at the data, we see that 36 patients had a disease recurrence within 10 months. Therefore,
the fraction of recurrence times less than or equal to 10 is 36/10 = 0.72.