4 - JustAnswer



4.) Marketing: age what is the age distribution of promotion-sensitive shoppers? A supermarket super shopper is defined as a shopper for whom at least 70% of the items purchased where on sale or purchased with a coupon. The following table is based on information taken form Trends in the United States (Food Marketing Institute, Washington, D.C.)

Age range, years 18-28 29-39 40-50 51-61 62 and over

Midpoint x 23 34 45 56 67

Percent of

Super shoppers 7% 44% 24% 14% 11%

For the 62 and over age group, use the midpoint 67 years.

a.) using the age midpoints x and the percentages of super shoppers, do we have a valid probability distribution? Explain.

Yes, we have a valid probability distribution. Probabilities of all possibilities add up to 100%.

b.) use a histogram to graph the probability distribution of part (a)

[pic]

c.) compute the expected age of μ of a super shopper.

[pic]

d.) compute the standard deviation σ for ages of super shoppers.

[pic]

6.) Sociology: Dress Habits A research team at Cornell University conducted a study sowing that approximately 10% of all businessmen who wear ties wear them so tight that they actually reduce blood flow to the brain, diminishing cerebral functions. At a board meeting 20 businessmen, all of whom wear ties, what is the probability that

a.) at least one tie is too tight?

P(X ≥ 1) = 1 - P(X = 0)

= 1 - 0.1216

= 0.8784

b.) more than two ties are too tight?

P(X > 2) = 1 - P(X ≤ 2)

= 1 - 0.6769

= 0.3231

c.) no tie is too tight?

P(X = 0) = 0.1216

d.) at least 18 ties are no too tight?

P(X ≥ 18)' = P(X ≤ 17)

≈ 1

4.) Insurance: Auto The Mountain States Office of State Farm Insurance Company reports that approximately 85% of all automobile damage liability claims were made by people under 25 years of age. A random sample of five automobile insurance liability claims is under study.

a.) Make a histogram showing the probabilities that r=0 to 5 claims are made by people under 25 years of age.

[pic]

b.) fin t he mean and the standard deviation of this probability distribution. For samples of size 5, what is the expected number of claims made by people under 25 years of age?

[pic] [pic]

Expected number of claims is 4.25

6.) Focus Problem: Personality Types We know have the tools to solve The Chapter Focus Problem. In the book A Guide to the Development and Use of the Myers-Briggs Type Indicators by Myers and McCaully, it was reported that approximately 45% of all university professors are extroverted. Suppose you have a class with six different professors

a.) what is the possibility that all six are extroverts?

P(X = 6) = 0.0083

b.) what is the possibility that none of your professors are extroverts?

P(X = 0) = 0.0277

c.) what is the probability that at least two of your professors are extroverts?

P(X ≥ 2) = 1 - P(X ≤ 1)

= 1 - 0.1636

= 0.8364

d.) in a group of six professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution?

E[X] = np = 6(0.45) = 2.7

s = [pic] = 1.219

2.) Random Variables: Classification which of the following are continuous variables and which are discrete?

a.) speed of an airplane

Continuous

b.) Age of a college professor

Discrete

6.) History: Florence Nightingale What was the age distribution of nurses in Great Britain at the time of Florence Nightingale? Thanks to Florence Nightingale and the British census of 1851, we have the following information. Note in 1851 there were 25,466 nurses in Great Britain, further more a strict distinction between nurses and domestic servants.

Age range (yr) 20-29 30-39 40-49 50-59 60-69 70-79 80+

Midpoint x 24.5 34.5 44.5 54.5 64.5 74.5 84.5

% of nurses 5.7% 9.7% 19.5% 29.2% 25.0% 9.1% 1.8%

a.) using the age midpoints x and the percentages of nurses, do we have a valid probability distribution? Explain.

Yes, we have a valid probability distribution. Probabilities of all possibilities add up to 100%.

b.) use a histogram to graph the probability distribution of part (a)

[pic]

c.) find the probabilities that a British nurse selected at random in 1851 would be 60 years of age or older.

P(X ≥ 60) = 25% + 9.1% + 1.8% = 35.9% = 0.359

d.) compute the expected age of µ of a British nurse contemporary to Florence nightingale

µ = 53.76

e.) compute the standard deviation of s for ages of nurses shown in the distribution.

s = 13.66

6.1

1.) Which of the following are continuous variables and which are discrete?

a.) number of traffic fatalities per year in the state of Florida

b.) distance golf ball travels after being hit with a driver

c.) time required to drive from home to college on any given day

d.) number of ships in Pearl Harbor on any given day

e.) your weight before breakfast each morning

a, and d are discrete

b, c, e are continuous

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download