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Chapter 1: Q21E (page 27)

The article cited in Exercise 20 also gave the following values of the variables y=number of culs-de-sac and z=number of intersections:

y

1

0

1

0

0

2

0

1

1

1

2

1

0

0

1

1

0

1

1

z

1

8

6

1

1

5

3

0

0

4

4

0

0

1

2

1

4

0

4

y

1

1

0

0

0

1

1

2

0

1

2

2

1

1

0

2

1

1

0

z

0

3

0

1

1

0

1

3

2

4

6

6

0

1

1

8

3

3

5

y

1

5

0

3

0

1

1

0

0

z

0

5

2

3

1

0

0

0

3

a. Construct a histogram for the ydata. What proportion of these subdivisions had no culs-de-sac? At least one cul-de-sac?

Short Answer

Expert verified

a.

The histogram is represented as,

Theproportion ofsubdivisions that had no culs-de-sac is 0.362.

The proportion of subdivisions that had at least oneculs-de-sac is 0.638.

Step by step solution

01

Given information

It is given that y represents the number of culs-de-sac and z represents the number of interactions.

02

Construct a histogram.

a.

The total number of culs-de-sac is 47.

The relative frequency is computed as,

\({\rm{relative frequency }} = \frac{{frequency}}{{Total\;number\;of\;observations}}\)

The table representing the relative frequency is computed as,

y

Frequency

Relative frequency

0

17

0.362

1

22

0.468

2

6

0.128

3

1

0.021

4

0

0.000

5

1

0.021

Steps to construct a histogram are,

1) Determine the frequency or the relative frequency.

2) Mark the class boundaries on the horizontal axis.

3) Draw a rectangle on the horizontal axis corresponding to the frequency or relative frequency.

The histogram is represented as,

03

Compute the proportion

The proportionof subdivisions that had no culs-de-sac is computed as,

\(P\left( {y = 0} \right) = 0.362\)

Therefore, theproportion ofsubdivisions that had no culs-de-sac is 0.362.

The proportionof subdivisions that had at least one culs-de-sac is computed as,

\(\begin{aligned}P\left( {y \ge 1} \right) &= 1 - P\left( {y = 0} \right)\\ &= 1 - 0.362\\ &= 0.638\end{aligned}\)

Therefore, the proportion of subdivisions that had at least one culs-de-sac is 0.638.

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Most popular questions from this chapter

Many universities and colleges have instituted supplemental

instruction (SI) programs, in which a student facilitator meets regularly with a small group of students enrolled in the course to promote discussion of course material and enhance subject mastery. Suppose that students in a large statistics course (what else?) are randomly divided into a control group that will not participate in SI and a treatment group that will participate. At the end of the term, each student鈥檚 total score in the course is determined.

a. Are the scores from the SI group a sample from an existing population? If so, what is it? If not, what is the relevant conceptual population?

b. What do you think is the advantage of randomly dividing the students into the two groups rather than letting each student choose which group to join?

c. Why didn鈥檛 the investigators put all students in the treatment group? [Note:The article 鈥淪upplemental Instruction: An Effective Component of Student Affairs Programming鈥� (J. of College Student Devel., 1997: 577鈥�586) discusses the analysis of data from several SI programs.]

A deficiency of the trace element selenium in the diet can negatively impact growth, immunity, muscle and neuromuscular function, and fertility. The introduction of selenium supplements to dairy cows is justified when pastures have low selenium levels. Authors of the article 鈥淓ffects of Short Term Supplementation with Selenised Yeast on Milk Production and Composition of Lactating Cows鈥� (Australian J. of Dairy Tech., 2004: 199鈥�203) supplied the following data on milk selenium concentration (mg/L) for a sample of cows given a selenium supplement and a control sample given no supplement, both initially and after a 9-day period.

Obs

InitSe

InitCont

FinalSe

FinalCont

1

11.4

9.1

138.3

9.3

2

9.6

8.7

104.0

8.8

3

10.1

9.7

96.4

8.8

4

8.5

10.8

89.0

10.1

5

10.3

10.9

88.0

9.6

6

10.6

10.6

103.8

8.6

7

11.8

10.1

147.3

10.4

8

9.8

12.3

97.1

12.4

9

10.9

8.8

172.6

9.3

10

10.3

10.4

146.3

9.5

11

10.2

10.9

99.0

8.4

12

11.4

10.4

122.3

8.7

13

9.2

11.6

103.0

12.5

14

10.6

10.9

117.8

9.1

15

10.8

121.5

16

8.2

93.0

a. Do the initial Se concentrations for the supplementand control samples appear to be similar? Use various techniques from this chapter to summarize thedata and answer the question posed.
b. Again use methods from this chapter to summarizethe data and then describe how the final Se concentration values in the treatment group differ fromthose in the control group.

Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be independent \({\rm{rv's}}\) with mean values \({{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}\) and variances \({\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{, \ldots ,\sigma }}_{{{\rm{\sigma }}^{\rm{*}}}}^{\rm{2}}\)Consider a function\({\rm{h}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}} \right)\), and use it to define a \({\rm{rv}}\)\({\rm{Y = h}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}} \right)\). Under rather general conditions on the \({\rm{h}}\)function, if the \({{\rm{\sigma }}_{\rm{i}}}\)'s are all small relative to the corresponding \({{\rm{\mu }}_{\rm{i}}}\)'s, it can be shown that (\({\rm{E(Y)\gg h}}\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}} \right)\) and\({\rm{V(Y)\hat U }}{\left( {\frac{{{\rm{露h }}}}{{{\rm{露 }}{{\rm{x}}_{\rm{1}}}}}} \right)^{\rm{2}}}{\rm{ \times \sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + L + }}{\left( {\frac{{{\rm{露h }}}}{{{\rm{露 }}{{\rm{x}}_{\rm{n}}}}}} \right)^{\rm{2}}}{\rm{ \times \sigma}}_{\rm{n}}^{\rm{2}}\))where each partial derivative is evaluated at \(\left( {{{\rm{x}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}} \right){\rm{ = }}\)\(\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}} \right)\). Suppose three resistors with resistances \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}\)are connected in parallel across a battery with voltage\({{\rm{X}}_{\rm{4}}}\). Then by Ohm's law, the current is

\({\rm{Y = }}{{\rm{X}}_{\rm{4}}}\left( {\frac{{\rm{1}}}{{{{\rm{X}}_{\rm{1}}}}}{\rm{ + }}\frac{{\rm{1}}}{{{{\rm{X}}_{\rm{2}}}}}{\rm{ + }}\frac{{\rm{1}}}{{{{\rm{X}}_{\rm{3}}}}}} \right)\)

Let \({{\rm{\mu }}_{\rm{1}}}{\rm{ = 10}}\)ohms, \({{\rm{\sigma }}_{\rm{1}}}{\rm{ = 1}}{\rm{.0ohm,}}\quad {{\rm{\mu }}_{\rm{2}}}{\rm{ = 15}}\)ohms, \({{\rm{\sigma }}_{\rm{2}}}{\rm{ = 1}}{\rm{.0ohm,}}{{\rm{\mu }}_{\rm{3}}}{\rm{ = 20}}\)ohms, \({{\rm{\sigma }}_{\rm{3}}}{\rm{ = 1}}{\rm{.5ohms,}}{{\rm{\mu }}_{\rm{4}}}{\rm{ = 120\;V}}\),

\({{\rm{\sigma }}_{\rm{4}}}{\rm{ = 4}}{\rm{.0 \backslash mathrm\{ \;V}}\)}. Calculate the approximate expected value and standard deviation of the current (suggested by "6andom Samplings," CHEMTECH, \({\rm{1984: 696 - 697}}\)).

a. \({\rm{Show}}\)that\({\rm{Cov(X,Y + Z) = Cov(X,Y) + Cov(X,Z}}\)).

b. Let \({{\rm{X}}_{\rm{1}}}\)and \({{\rm{X}}_{\rm{2}}}\)be quantitative and verbal scores on one aptitude exam, and let \({{\rm{Y}}_{\rm{1}}}\)and \({{\rm{Y}}_{\rm{2}}}\)be corresponding scores on another exam. If\({\rm{Cov}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{Y}}_{\rm{1}}}} \right){\rm{ = 5}}\), \({\rm{Cov}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{Y}}_{\rm{2}}}} \right){\rm{ = 1,Cov}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{Y}}_{\rm{1}}}} \right){\rm{ = 2}}\), and \({\rm{Cov}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{Y}}_{\rm{2}}}} \right){\rm{ = 8}}\), what is the covariance between the two total scores \({{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\)and\({{\rm{Y}}_{\rm{1}}}{\rm{ + }}{{\rm{Y}}_{\rm{2}}}\)?

Each customer making a particular Internet purchase must pay with one of three types of credit cards (think Visa, MasterCard, AmEx). Let \({{\rm{A}}_{\rm{i}}}{\rm{(i = 1,2,3)}}\) be the event that a type \({\rm{i}}\)credit card is used, with\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = }}{\rm{.5}}\), \({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = }}{\rm{.3}}\), and\({\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}{\rm{.2}}\). Suppose that the number of customers who make such a purchase on a given day is a Poisson \({\rm{rv}}\) with parameter\({\rm{\lambda }}\). Define \({\rm{rv}}\) 's \({{\rm{X}}_{{\rm{1,}}}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}\)by \({{\rm{X}}_{\rm{i}}}{\rm{ = }}\)the number among the \({\rm{N}}\)customers who use a type \({\rm{i}}\)card\({\rm{(i = 1,2,3)}}\). Show that these three \({\rm{rv}}\) 's are independent with Poisson distributions having parameters\({\rm{.5\lambda ,}}{\rm{.3\lambda }}\), and\({\rm{.2\lambda }}\), respectively. (Hint: For non-negative integers\({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}\), let\({\rm{n = }}{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}\). Then\({\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}} \right.\), \(\left. {{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}} \right){\rm{ = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}{\rm{,N = n}}} \right)\) (why is this?). Now condition on\({\rm{N = n}}\), in which case the three \({\rm{Xi}}\) 's have a trinomial distribution (multinomial with three categories) with category probabilities\({\rm{.5,}}{\rm{.3}}\), and . \({\rm{2}}\).)

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