BUS 350 HW 6

Airlines sometimes overbook flights (that is, they sell more tickets than there are seats on the plane). Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable X as the number of ticketed passengers who actually show up for the flight. The probability distribution function of X is shown in the table below.

x

45

46

47

48

49

50

51

52

53

54

55

p(x)

0.05

0.10

0.12

0.14

0.25

0.17

0.06

0.05

0.03

0.02

0.01

Question 1. What is the probability that the flight will accommodate all ticketed passengers who show up? .83

0.83

Question 2. Calculate E(X), the expected number of ticketed passengers who show up for the flight.

48.84

48.8

Question 3. If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? .66

0.66

Question 4. If you are the third person on the standby list, what is the probability that you will be able to take the flight?

.27

.27

  1. 2.5/2.5 points | Previous Answers

You are planning to take to a trip to Montreal, Canada during the month of April and you want to bring clothing that is appropriate for the weather. The daily high temperature X in degrees Celsius in Montreal during April has expected value E(X) = 9.6oC with a standard deviation SD(X) = 4oC. You want to convert these Celsius temperatures to oF (degrees Fahrenheit). The conversion of X into degrees Fahrenheit Y is Y = (9/5)X + 32.

Question 1. What is E(Y), the expected daily high in Montreal during April in degrees Fahrenheit?

49.28

49.3

Question 2. What is SD(Y), the standard deviation of the daily high temperature in Montreal during April in degrees Fahrenheit?

7.2

7.2

Solution or Explanation

Question 1: E(Y) = (9/5)*E(X)+32

Question 2: SD(Y) = (9/5)*SD(X)

  1. –/2 points

Recent data show that women's ice hockey is the most dangerous NCAA sport in terms of concu hockey and even football (data from NYTimes and NCAA). This result is even more surprising w allowed in women's ice hockey.

The first table below shows the probability distribution of the number of concussions per 1,000 hockey. The second table shows the probability distribution for NCAA men's ice hockey.

NCAA Women's Ice Hockey

Number of Concussions per 1,000 player hours

0

1

2

3

4

5

6

7

8

Probability

0.06587

0.17918

0.24368

0.22094

0.15024

0.08173

0.03705

0.01440

0.004

NCAA Men's Ice Hockey

Number of Concussions per 1,000 player hours

0

1

2

3

4

5

Probability

0.22993

0.33799

0.24842

0.12173

0.04473

0.013

Buffy and her boyfriend Bubba both play college hockey at the same university.

Question. Over 1,000 player hours what is the probability that the number of concussions on B deviation of the mean? (mean is another name for the expected value)

Answer the same question for Bubba's team.

Note: calculate the means and standard deviations to 2 decimal places. (No Response) 0.61486

probability for Buffy's team

0.58641

(No Response) probability for Bubba's team

  1. 4/4 points | Previous Answers

On Valentine's Day the Quiet Nook restaurant offers a Lucky Lovers Special that could save couples money on their romantic dinners. When the waiter brings the check he also brings a scratch-off ticket that the couple scrapes to reveal a hidden number; call this number X. The value of the number X is the dollar amount that will be subtracted from their dinner bill. The possible values of the scratch-off number X are $0, $10, $20, $25, and $30. The table below shows the probability distribution of the values of X.

Discount Value X

$0

$10

$20

$25

$30

Probability

0.25

0.30

0.25

0.10

0.10

  1. Find the expected value of the discount X.$ 5

13.50

  1. Find the standard deviation of the discount X.$ 01

10.01

(Use 2 decimal places.)

For the past several weeks the restaurant has also been distributing coupons worth $5 off any dinner for two. The coupon can be used in addition to the discount X from the scratch-off ticket. If every couple dining there on Valentine's Day also brings a $5 coupon, what are the expected value and standard deviation of the total discount the couples receive?

  1. Expected value of total discount.$ 5

18.50

  1. Standard deviation of total discount$ 01

10.01

When two couples dine together on a single check, the restaurant doubles the scratch-off discount X; so the discount values X are $0, $20, $40, $50, and $60 (the $5 discount coupon cannot be used).

  1. What is the expected value of the discount for 2 couples?

$ 27

27.00

  1. What is the standard deviation of the discount for 2 couples?$ 02

20.02

Solution or Explanation

Question a: E(X) = 540*(0.1)+545*(0.25)+550*(0.3)+555*(0.25)+560*(0.1) = 550

SD(X) = sqrt[(540-550)2*(0.1)+(545-550)2*(0.25) +(550-550)2*(0.3)+(555-550)2*(0.25) + (560-550)2*(0.1)] = 5.701

Question c: E(X-550) = E(X)-550 = 550-550 = 0

Question d: SD(X-550) = SD(X) = 5.701

  1. 5/3.5 points | Previous Answers

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9 because we do not write numbers such as 15 as 015. While it is reasonable to think that for most real-life data each digit occurs with equal frequency so that each digit has probability 1/9 of being the first significant digit, this is not true. It is a surprising phenomenon that in many naturally occurring numbers and web-based data the first significant digit has a probability distribution known as Benford's law.

From Wikipedia:

Benford's law, also called the first-digit law, states that in lists of numbers from many (but not all) real-life sources of data, the leading digit is distributed in a specific, non-uniform way.

Specifically, for d = 1, 2, 3, 4, 5, 6, 7, 8, 9, Benford's law states P(first significant digit is d) = log_10(1+(1/d)). The distribution of the first digit according to Benford's law, calculated to 3 decimal places, is shown in the table below.

Benford's law

First Digit X

1

2

3

4

5

6

7

8

9

Probability

.301

.176

.125

.097

.079

.067

.058

.051

.046

Benford's Law and the Equally Likely Model

The law is named after physicist Frank Benford, who stated it in 1938, although it had been previously stated by Simon Newcomb in 1881. A surprising variety of data from the natural sciences, social affairs, and business obeys Benford's law. Quoting Wikipedia again:

This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature).

Numbers that are assigned, such as social security numbers and zip codes, or data with a fixed maximum, such as deductible contributions to individual retirement accounts, or randomly generated numbers, do not follow Benford's law.

The figure below shows how the distribution of the first digit of various naturally occurring and web-based data compares with Benford's law.

Figure information.Earthquakes: depth in km of 248,915 quakes, 1989-2009; source National Earthquake Information Center, United States Geological Survey. Minnesota lakes: size in acres of approx. 1100 lakes; source Wikipedia. Births: number of births in each county of the United States (approximately 3200), 2010; source: US Census Bureau. Diggs: total number of diggs for each of the top 1000 diggers at digg.com; source: socialblade.com.

Question 1:

(1a). What is the expected value of the first digit when the first digit follows Benford's law?

3.441 3.441 expected value (Use 3 decimal places).

(1b). What is the expected value of the first digit when the possible first digits are equally likely?

5 5

expected value (Use 3 decimal places).

(1c). What is the standard deviation of the first digit when the first digit follows Benford's Law?

2.462

2.462 standard deviation (Use 3 decimal places).

The tendency of many types of financial records to follow Benford's law provides a useful fraud detection tool in the field of financial forensics. An auditor or fraud investigator can, for example, compare first digits of invoices from a company's vendors with Benford's Law. A vendor whose invoices show a different pattern can be investigated for any "Enron-type" activities.

As an example of financial forensics, in State of Arizona v. Wayne James Nelson, Mr. Nelson was accused of trying to defraud the state of nearly $2 million. Nelson, a manager in the office of the Arizona State Treasurer, argued that he wasn't trying to steal the money but had diverted actual vendor invoice amounts to a bogus vendor only to demonstrate the absence of safeguards in a new computer system. The amounts of the 23 checks issued by Mr. Nelson are shown in this Excel file.

Question 2. What proportion of the checks had a leading digit of 7, 8, or 9? .913

0.913

(use 3 decimal places)

Question 3. Under Benford's law what is the probability of a leading digit of 7, 8, or 9? .155

0.155

(use 3 decimal places)

Question 4. You are the judge and you must focus on the facts to issue a verdict. As the judge, what would you decide?

  1. 3/3 points | Previous Answers

Investors, be they large corporations, banks, pension funds, mutual funds, or individuals, seldom hold a single financial asset; rather, they hold portfolios of financial assets. Thus, the investor should be less concerned with the rate of return achieved by, say a particular stock in the portfolio than the overall rate of return of the portfolio.

Typically, an investor can choose from among many different assets to form a portfolio; in other words, there are many different portfolios from which an investor can choose. But which portfolio should the investor select? Since the future rate of return of a portfolio is uncertain a probability distribution can be used to characterize a portfolio's future rate of return. In particular, investment managers frequently characterize portfolios by the mean mu (expected value) and standard deviation sigma (risk) of their rates of return. First, the set of all possible portfolios - - the feasible set - - must be reduced to an efficient set of portfolios. An efficient portfolio is one that provides the highest possible mean rate of return for any given degree of risk (i.e., any given standard deviation sigma) or the lowest possible degree of risk (i.e., the lowest standard deviation sigma) for any given mean rate of return.

The attached graph (right click here and choose "Open Link in New Window" to see graph) shows the mean and standard deviation of the rate of return for various portfolios. Portfolios identified by a value of mu and sigma that fall within the ellipse represent the feasible set of portfolios for a particular investor. The efficient set of portfolios is denoted by the boundary arc ABC and is sometimes called the efficient frontier. Portfolios to the left of ABC are not attainable because they fall outside the feasible set. Portfolios to the right of ABC are not efficient because there always exists a portfolio that could provide (1) a higher mean return for a given level of standard deviation of returns (compare points B and D), or (2) a lower standard deviation (lower degree of risk) for a given mean rate of return (compare points B and E which have the same mean rate of return).

The future rates of return of stock portfolios X, Y, and Z can be characterized by the following probability distributions:

STOCK PORTFOLIO X

Rate of Return

Probability

.20

.05

.15

.15

.10

.26

.05

.20

.00

.15

-.05

.10

-.10

.05

-.15

.03

-.20

.01

STOCK PORTFOLIO Y

Rate of Return

Probability

.15

.10

.10

.20

.05

.30

.00

.20

-.05

.10

-.10

.07

-.15

.03

STOCK PORTFOLIO Z

Rate of Return

Probability

.25

.05

.20

.10

.15

.25

.10

.20

.05

.15

.00

.10

-.05

.07

-.10

.05

-.15

.03

.20

  1. What is the probability of a negative rate of return if you invest in portfolio Y?

.20

  1. Find the expected rate of return of portfolio X. .052

.052

.

  1. Find the expected rate of return of portfolio Y. .0335

.0335

.

  1. Find the expected rate of return of portfolio Z. .0845

.0845

.

Suppose you plot the mean and standard deviation for each of the portfolios X, Y, and Z on a graph like the one in the above link. Note that the standard deviation of the rate of return is on the horizontal axis and the expected rate of return is on the vertical axis. (DO NOT CALCULATE THE STANDARD DEVIATIONS! The standard deviations are as follows: portfolio X: .087155; portfolio Y: .074181; portfolio Z: .0966).

  1. Below are statements about portfolios X, Y, and Z. Choose all the statements that are true. (This is a multiple-select problem; you must choose all the true statements to get the problem correct).

A young investor who wants more return and who is not very concerned about risk should choose portfolio Y over portfolio X.

An investor close to retirement who does not want to put her retirement nest egg at risk should invest in portfolio Z.

Portfolio Y is for timid investors who want the least amount of risk.

If you want to maximize your expected rate of return and are not concerned about risk, you would choose portfolio Z.

Portfolio X has more risk than portfolio Y.

The overall pattern in the plot of the three portfolios is from lower left to upper right, meaning that if you want a higher return you have to take more risk.

The overall pattern in the plot of the three portfolios is from lower left to upper right, meaning that portfolios with low risk have a high expected return.

  1. 1.5/1.5 points | Previous Answers

In the 4x100 medley relay event in swimming, four swimmers swim 100 yards, each using a different stroke. The four medley relay swimmers for a college team preparing for their conference championships have the following summary statistics (in seconds) for the four strokes:

Swimmer

Mean

St. Dev.

1 (backstroke)

50.72

0.21

2 (breaststroke)

55.51

0.33

3 (butterfly)

49.43

0.25

4 (freestyle)

44.91

0.22

Question. Let the random variable T denote the relay team's total time in the medley event. Determine the mean E(T) and standard deviation SD(T). 200.57 200.57

mean

0.514

.514 standard deviation (use 3 decimal places)

Solution or Explanation

E(T)=50.72+55.51+49.43+44.91=200.57 SD(T)=

sqrt(Var(backstroke)+Var(breaststroke)+Var(butterfly)+Var(freestyle))

  1. 4/4 points | Previous Answers

A "buy stock option" is a privilege sold by one party to another that gives the purchaser the right, but not the obligation, to buy a particular stock at an agreed-upon price on a specific date.

You have an opportunity to purchase a "buy stock option" for $200. If the stock closes above $30 on Dec. 15, the option will be worth $1000. If the stock closes below $20 on Dec. 15, the option will be worth nothing. If it closes between $20 and $30 inclusive, on Dec. 15, the option will be worth $200.

Your financial advisor tells you that there is a 50% chance that the stock will close in the $20 $30 range, a 20% chance that it will close above $30, and a 30% chance that the stock will close below $20 on Dec. 15.

Question 1. Which table below shows the probability distribution for the value of the stock option on Dec. 15.

Stock Option Value x

under $20

$20 to $30, inclusive

over $30

p(x)

.30

.50

.20

Stock Option Value x

$0

$200

$1000

p(x)

.50

.30

.20

Stock Option Value x

under $20

$20 to $30, inclusive

over $30

p(x)

.30

.20

.50

Stock Option Value x

$0

$200

$1000

p(x)

.30

.50

.20

Stock Option Value x

$0

$200

$1000

p(x)

.30

.30

.20

Question 2. How much do you expect the stock option to be worth? (DO NOT use a dollar sign in your answer). 300

300

Question 3. What is the standard deviation of the stock option's worth? (DO NOT use a dollar sign in your answer). 360.55 360.56

Question 4. How much do you expect to gain? (DO NOT use a dollar sign in your answer).

100

100

Question 5. What is the standard deviation of your gain? (DO NOT use a dollar sign in your answer).

360.55

360.56

.

Solution or Explanation

Question 2.

Let X = worth of stock option. E(X)=$0*(0.30)+$200(0.50)+$1000*(0.20) = $300.

Question 3.

VAR(X) = (0-300)2*(0.30)+(200-300)2*(0.50)+(1000-300)2*(0.20) = 130,000.

SD(X) = sqrt(130,000) = 360.56

Question 4.

Since you paid $200 for the stock option, your gain is X-200. E(X-200) = E(X)-200 = 300-200 = 100.

Question 5.

SD(X-200) = SD(X) = 360.56.

  1. 3/3 points | Previous Answers

To compete with Netflix, the owner of a local movie rental store decides to try sending DVDs through the mail. To plan for this endeavor, she sends DVDs to her friends to obtain data on delivery times when DVDs are mailed to customers and return times when DVDs are mailed back to the store.

Let the random variable X denote the delivery time when a DVD is mailed to a customer, and let the random variable Y denote the return time when a customer mails the DVD back to the store.

The store owner found that the mean delivery time, E(X), was 1.7 days, and that the standard deviation of delivery time, SD(X), is 0.3 days. She found that the mean return time, E(Y), is 2.5 days, with a standard deviation SD(Y) of 0.3 days.

Question 1. Determine the mean and standard deviation of the total transit time for a DVD (that is, the delivery time when mailed to the customer plus the return time when mailed back to the store).

4.2 4.2

days, mean total transit time

0.424

.424 days, standard deviation of total transit time (use 3 decimal places)

Question 2. On average the return time Y is greater than the delivery time X. Determine the mean and standard deviation of how much the return time Y exceeds the delivery time X. .8 0.8

days, mean of how much the return time exceeds the delivery time.

0.424

.424 days, standard deviation of how much the return time exceeds the delivery time.

Solution or Explanation

Question 1: E(X+Y) = E(X) + E(Y)

SD(X+Y) = sqrt[Var(X+Y)] = sqrt[Var(X) + Var(Y)]

Question 2: E(Y-X) = E(Y) - E(X)

SD(Y-X) = sqrt[Var(Y-X)] = sqrt[Var(Y) + Var(X)]

10.3/3 points | Previous Answers

A life insurance company sells a term insurance policy to a 21-year-old male that pays $100,000 if the insured dies within the next 5 years. The company collects a premium of $250 each year as payment for the insurance. The amount X that the company earns on this policy is $250 per year, less the $100,000 that it must pay if the insured dies. From mortality tables, the expected value of X, denoted E(X) or μ, is $303 and the standard deviation of X, denoted SD(X) or σ, is $9707.

Question 1. The risk of insuring one person's life is reduced if we insure many people. Suppose an insurance company insures two 21-year-old males and that their ages at death are independent. If X and Y are what the company earns from the two insurance policies, the insurance company's average income on the two policies is Z = (X+Y)/2 = 0.5X + 0.5Y.

Find the expected value E(Z) and the standard deviation SD(Z) of the random variable Z.

(Note: since the ages at death are independent, Var(X+Y)=Var(X)+Var(Y)) E(Z) = $ 303

303

6863.89

SD(Z) = $ 6864.29 .

Question 2. If four 21-year-old males are insured, the insurance company's average income is Z = 1/4(X_1 + X_2 + X_3 + X_4) where Xi is what the insurance company earns by insuring one man. The Xi are independent and each has the same distribution with expected value E(X) and standard deviation SD(X) as given above.

Find the expected value E(Z) and the standard deviation SD(Z) of the random variable Z. (Note: since the Xi's are independent, the variance of the sum of the Xi's is the sum of the individual variances). E(Z) = $ 303

303

4853.5

SD(Z) = $ 4853.78 .

Solution or Explanation Question 1:

.

therefore,

Question 2:

.

therefore,

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