a) Find the mean, median, and mode for each data set.
b) Which of these three measures is the weakest indicator of a "typical" data value?

(a)
100m dash times:

Mean=9.87+9.98+10.02+10.15+10.36+10.366=10.12Mean=9.87+9.98+10.02+10.15+10.36+10.366=10.12
Median=10.02+10.152=10.085Median=10.02+10.152=10.085
Mode = 10.36

Number of children:

Mean=0+1+1+2+2+2+2+2+2+2+2+2+613=2Mean=0+1+1+2+2+2+2+2+2+2+2+2+613=2
Mode = 2
Median = 2

Number of cars in driveway:

Mean=0+0+1+1+2+2+3+58=1.75Mean=0+0+1+1+2+2+3+58=1.75
Median=1+22=1.5Median=1+22=1.5
Mode = 0,1, and 2

(b)
100m dash times:

The mode is the weakest because all of the values fall at or below 10.36. We also know that the mode should be used for a small range of discrete data or attribute data. This is a small range of continuous data.

Number of children:
All measures of central tendency in this case have the same value of 2, which is a strong indicator of a "typical" data value.

Number of cars in driveway:
The mode is the weakest because there are three different values for mode and only five unique values in the entire data set.

Which type of data (cross-sectional or time series) is each variable?

a) Mexico's GDP for each of the last 10 quarters.
b) Unemployment rates in each of the 31 states in Mexico at the end of last year.
c) Unemployment rate in Mexico at the end of each of the last 10 years.
d) Average home value in each of the 10 largest Mexican cities today.

a) Time series
b) Cross-sectional
c) Time series
d) Cross-sectional

The target population is all stocks in the S&P 500 index. Is each of the following a parameter or a statistic?

a) The average price/earnings ratio for all 500 stocks in the S&P index.
b) The proportion of all stocks in the S&P 500 index that had negative earnings last year.
c) The proportion of energy-related stocks in a random sample of 50 stocks.
d) The average rate of return for 20 stocks recommended by a broker.

Explanation

(a) Parameter. The S&P is the population.

(b) Parameter. Same as above. The S&P is the population.

(c) Statistic. We clearly stated a random sample.

(d) Statistic. This isn’t random, but it could be considered a sample.

Would you use a sample or a census to measure each of the following?

a) The number of cans of Campbell's soup on your local supermarket's shelf today at 6:00 p.m.
b) The proportion of soup sales last week in Boston that was sold under the Campbell's brand.
c) The proportion of Campbell's brand soup cans in your family's pantry.

Explanation

(a) Census. It would be easy enough to count all of them.

(b) Sample. It would be too costly to track each can.

(c) Census. You can count them all quickly and cheaply.

For each data set, is the mode a good measure of center?

a) Genders of 12 CEOs: M, M, F, M, F, M, M, M, F, M, M, M
b) Ages of 10 college freshmen: 17, 17, 18, 18, 18, 18, 18, 18, 19, 20
c) Ages of 8 MBA students: 24, 26, 27, 28, 30, 31, 33, 37

Explanation

a. Yes
This is attribute data so mode is the only measure of central tendency possible. Mode = M, which occurs 9 out of 12 observations.

b. Yes
This is discrete data with a small range so mode could be an appropriate measure of central tendency. Mode = 18, which occurs 6 out of 10 observations.

c. No
There is no mode in this case because there is no value that occurs more than once. Therefore, either mean or median would be more appropriate. Mean = 29.5 and median = 29.

A certain health maintenance organization (HMO) is studying its daily office routine. It collects information on three variables: the number of patients who visit during a day, the patient’s complaint, and the waiting time until each patient sees a doctor.

a) Which variable is categorical?
- The number of patients who visit during a day
- The waiting time until each patient sees a doctor
- The patient's complaint
b) Identify the two quantitative variables and state whether they are discrete or continuous.

a) The patient's complaint
b) The number of patient visits is discrete numerical. The waiting time is continuous.

Consider the following table of data.

a) Select the correct chart to display the above data
b) Would more than one kind of display be acceptable for this data?

- Yes, a bar chart or column chart would also work.
- No, other charts would not work for this data.
- Yes, a line chart would also work.

a) Chart A
b) Yes, a bar chart or column chart would also work.

One disadvantage of the range is that

a) it is expressed in different units than the mean.
b) it is undefined if any X values are 0 or negative.
c) it does not exist for some data sets.
d) only extreme values are used in its calculation.

d) only extreme values are used in its calculation.

Explanation
Range ignores all the sample except the extremes x _max and x _min.

Noodles and Company tested consumer reaction to two spaghetti Sauces. Each of 70 raters assessed both Sauces on a scale of 1 (worst) to 10 (best) using several taste criteria. To correct for possible bias in tasting order, half the raters tasted Sauce A first, while the other half tasted Sauce B first. Actual results are shown below for “overall liking.”

Sauce A:
6, 7, 7, 8, 8, 6, 8, 6, 8, 7, 8, 8, 6, 8, 7, 7, 7, 8, 8, 8, 7, 7, 6, 7, 7,
8, 3, 8, 8, 7, 8, 6, 7, 8, 7, 7, 3, 6, 8, 7, 1, 8, 8, 7, 6, 7, 7, 4, 8, 8,
3, 8, 7, 7, 7, 5, 7, 7, 7, 9, 5, 7, 6, 8, 8, 8, 4, 5, 9, 8

Sauce B:
7, 7, 7, 8, 8, 7, 8, 6, 8, 7, 7, 6, 7, 7, 8, 7, 8, 7, 8, 8, 7, 8, 5, 7, 7,
9, 4, 8, 8, 7, 8, 8, 8, 8, 7, 7, 3, 7, 9, 8, 9, 7, 8, 8, 6, 7, 7, 7, 8, 8,
7, 7, 8, 6, 6, 7, 7, 9, 7, 9, 8, 8, 6, 7, 7, 9, 4, 4, 9, 8

a) Calculate the mean and standard deviation for each sample.
b) Calculate the coefficient of variation for each sample.
c) What is your conclusion about consumer preferences for the two Sauces?
- On average, consumers seem to prefer Sauce A over Sauce B.
- On average, consumers seem to prefer Sauce B over Sauce A.

c) On average, consumers seem to prefer Sauce B over Sauce A.

Explanation

(a)
x¯A=6.86,SA=1.497,x¯B=7.24,SB=1.209 x¯A=6.86,SA=1.497,x¯B=7.24,SB=1.209 ⁢
(b)
CVA =1.497/6.86=0.218 or  21.8%,CVB  =1.209/7.24  = 0.167 or  16.7%CVA⁢ =1.497/6.86=0.218⁢ or⁢  21.8%,CVB⁢ ⁢ =1.209/7.24⁢  = 0.167⁢ or⁢  16.7%

(c)
Consumers preferred Sauce B. Sauce B also had more consistent ratings relative to the average.

Which distribution is least likely to be skewed to the right by high values?

a) Annual incomes of n passengers on a flight from New York to London
b) Weekend gambling losses of n customers at a major casino
c) Cost of a plain McDonald’s hamburger in n U.S. cities
d) Accident damage losses by n renters of an auto rental company

c) Cost of a plain McDonald’s hamburger in n U.S. cities

Explanation
A few high values would skew the data badly in all but the hamburger example, because a McDonald’s hamburger is a standard menu item.

Consider the following sample data:

Sample A: 6, 7, 8
Sample B: 61, 62, 63
Sample C: 1,000; 1,001; 1,002

a) Find the mean and standard deviation for each sample.
b) What does this exercise show about the standard deviation?
- The idea is to illustrate that the standard deviation is not a function of the
value of the mean.
- The idea is to illustrate that the standard deviation is a function of the value
of the mean.

a) Sample A: Sample B: Sample C:
Mean = 7 62 1,001
Sample Std Deviation = 1 1 1

b) The idea is to illustrate that the standard deviation is not a function of the value of the mean.

Explanation
(b)
The midpoint of each sample is the mean. The other 2 data points are exactly 1 standard deviation from the mean. The idea is to illustrate that the standard deviation is not a function of the value of the mean.

The asset turnover ratio (ATR) is the ratio of a company’s revenues to the value of its assets (indicating its efficiency in deploying its assets). We should not use the standard deviation to compare ATR variation among industrial sectors because firms with large asset bases (e.g., utilities, financial) typically have lower mean ATR than, say, retail firms.

a) Use the sample data to calculate the coefficient of variation for each sector. (Round your answers to 2 decimal places.)
b) Which sector has the highest degree of relative variation? The lowest?
- The _______ sector has the highest relative variation and ______ has the lowest relative variation.
c) If someone (incorrectly) used the standard deviations to compare variation, would the ranking among sectors be the same?

Explanation
(a)
The ATR coefficient of variation for Retail: 23.18%, for Utilities: 18.82%, for Financial: 39.97%
(b) The financial sector has the highest relative variation and utilities has the lowest relative variation.
(c)
If one were to compare standard deviation instead of CV then it would appear that the retail sector has the highest variability.

Which survey questions below would be suitable for a set of check boxes providing choices for the respondent to choose from? (You may select more than one answer. Click the box with a check mark for the correct answer and double click to empty the box for the wrong answer.)

Check All That Apply

a) How concerned are you for the environment?
b) What company do you work for?
c) What is your salary range?
d) What was the amount of your last utility bill?

a) How concerned are you for the environment?

c) What is your salary range?

Explanation
When asking about your level of concern for the environment, possible choices include “Very,” “Moderately,” or “Not at all.” Likewise, when asking what your salary range is, possible choices include “$0 to $25,000,” “$25,001 to $40,000,” or “$40,001 to $75,000.”

What type of data (categorical, discrete numerical, or continuous numerical) is each of the following variables?

a) The manufacturer of your car.
b) Your college major.
c) The number of college credits you are taking.

a) Categorical
b) Categorical
c) Discrete Numerical

The table shows the number of visits to Skier/Snowboarder from 1992 to 2016 in US.

a) Select the line chart of the Skier/snowboarder data.
b) Describe the pattern, if any.
- The number of visits showed a steady decrease until a sharp high in 2010-
2011 season.
- The number of visits showed a steady increase until a sharp drop in 2011-
2012 season.
- The number of visits showed a steady increase until a sharp drop in 2005-
2006 season.

a) Chart A
b)The number of visits showed a steady increase until a sharp drop in 2011-2012 season.

Explanation
(b)
In spite of variation from year to year the number of visits showed a steady increase until a sharp drop in 2011-2012 season.

For each data set, find the mean, median, and mode.

a) Campus health center visits (12 students): 0, 0, 0, 0, 0, 1, 2, 3, 3, 5, 5, 15 (Round your answers to 2 decimal places.)

b) Red Rocks ticket prices (9 concerts): 40, 40, 65, 71, 72, 75, 76, 78, 98 (Round your answers to 2 decimal places.)

c) Sodium grams in canned soup (8 varieties): 225, 255, 295, 302, 304, 337, 351, 366 (Round your answers to 2 decimal places.)

Explanation
(a)
Mean = 2.83, Median = 1.5, Mode = 0. Any one of these would be a useful measure of center. Data appears skewed right with many values equal to zero so one might choose median or mode over mean.
(b)
Mean = 68.33, Median = 72, Mode = 40. This is continuous numerical data so either the mean or the median would be best.
(c)
Mean = 304.38, Median = 303, No mode. This is continuous numerical data so either the mean or the median would be best.

a) Select the correct 2-D pie chart for these web-surfing data.
b) Select the correct exploded 2-D pie chart.
c) Select the correct bar chart.

a) Chart A
b) Chart F
c) Chart I

As a measure of variability, compared to the range, an advantage of the standard deviation is that it:

a) describes the distance between the highest and lowest values.
b) considers all data values.
c) considers only the data values in the middle of the data array.
d) is calculated easily through the use of a formula.

b) considers all data values.

Explanation
The range is easy to calculate but utilizes only two data values, which may be unusual.

Five homes were recently sold in Oxnard Acres. Four of the homes sold for $400,000, while the fifth home sold for $2.5 million. Which measure of central tendency best represents a typical home price in Oxnard Acres?

a) The midrange or mean.
b) The mean or mode.
c) The median or mode.
d) The mean or median.

c) The median or mode.

Explanation
Outliers will affect the mean or midrange.

What type of data (categorical, discrete numerical, or continuous numerical) is each of the following variables?

a) Length of a TV commercial.
b) Number of peanuts in a can of Planter's Mixed Nuts.
c) Occupation of a mortgage applicant.
d) Flight time from London Heathrow to Chicago O'Hare.

a) Continuous Numerical
b) Discrete Numerical
c) Categorical
d) Continuous Numerical

Analysis of portfolio returns over a 20-year period shows the statistics below.

a) Calculate and compare the coefficients of variation. (Round your answers to 2 decimal places.)
b) Why would we use a coefficient of variation, and why not just compare the standard deviations?
- The standard deviations are an "absolute", not relative, measure of dispersion.
It is best to use the CV when comparing across variables that have different
means
- Standard deviation can only be compared when the variables have different
units of measure.
- The standard deviations are relative and not absolute measures of dispersion.
c) What do the data tell you about risk and return at that time period?
- Venture funds have greater risk and lower return than common stocks based
on the CV.
- Venture funds have lower risk and greater return than common stocks based
on the CV.
- Federal short-term paper has the lowest standard deviation and hence the
greatest risk; real estate, the lowest risk.
- Federal short-term paper has the lowest CV and hence the greatest risk; real
estate, the lowest risk.

Explanation
(a) Answers in image
(b)
The standard deviations are “absolute”, not relative, measures of dispersion. It is best to use the CV when comparing across variables that have different means.
(c)
The risk and returns are captured by the CV. Federal short term paper has the lowest CV and hence lowest risk, Real estate the greatest risk. Venture funds have lower risk and greater return than common stocks based on the CV. In other words, there is more risk when there is more variation in the returns.

Would you use a sample or census to measure each of the following?
a) The number of workers currently employed by Campbell Soup Company.
b) The average price of a can of Campbell's Cream of Mushroom soup.
c) The total earnings of workers employed by Campbell Soup Company last year.

Explanation
(a) Census. This is assuming the company can easily generate the value from its human resource center.
(b) Sample. Impossible to observe prices of all cans in grocery stores.
(c) Census. This should be in Campbell Soup's data base

Below is a frequency distribution of earnings of 50 contractors in a country.

Regarding this distribution, which of the following is correct?

a) The frequency distribution contains too many class intervals.
b) The class intervals are too wide.
c) The class interval limits are ambiguous.
d) Too few classes were chosen.

c) The class interval limits are ambiguous.

Explanation
There are overlapping bin limits.

Days on the market are shown for the 36 most recent home sales in the city of Sonando Hills.

a) Calculate the mean, median, and mode. (Round your answers to 2 decimal places.)
b) Is the distribution skewed?
- Yes, the distribution is skewed right.
- No, the distribution is symmetric
- Yes, the distribution is skewed left.
c) Is the mode a useful measure of center for this data set?
- Yes
- No

a) Mean = 98.53
Median = 98.50
Mode = 20
b) Yes, the distribution is skewed right.
c) No

Is each of the following a parameter or statistic?

a) The number of visits to a pediatrician’s office last week.
b) The number of copies of John Grisham’s most recent novel sold to date.
c) The total revenue realized from sales of John Grisham’s most recent novel.

Explanation

(a) Statistic. The week of visits is the sample.
(b) Parameter. The population is all books sold to date.
(c) Parameter. The population is all books sold.

The table shows the number of visits to Skier/Snowboarder from 1992 to 2016 in US.

a) Select the line chart of the skier/snowboarder data.
b) Describe the pattern, if any.
- a steady increase until a sharp drop in 2014-2015 season.
- The number of visits showed a steady increase until a sharp drop in 2002-
2003 season.
- The number of visits showed a steady decrease until a sharp high in 2014-
2015 season.

Explanation
(a)
Chart A
(b)
In spite of variation from year to year the number of visits showed a steady increase until a sharp drop in 2014-2015 season.

Identify the following data as either time series or cross-sectional.
a) The 2017 CEO compensation of the 500 largest U.S. companies.
b) The annual compensation for the CEO of Coca-Cola Enterprises from 2010 to 2017.
c) The weekly revenue for a Noodles & Company restaurant for the 52 weeks in 2017.
d) The number of skiers on the mountain on Christmas Day 2017 at each of the ski mountains owned by Vail Resorts.

Explanation

(a) Cross-sectional. A single point in time: end of 2017.
(b) Time series. Data are collected over an 8-year time period.
(c) Time series. Data collected over 52 weeks.
(d) Cross-sectional. Single point in time: Christmas Day 2017

Three randomly chosen Colorado students were asked how many times they went rock climbing last month. Their replies were 5, 6, 7. The standard deviation is

a) 1.212
b) 1.000
c) 0.816
d) 1.056

b) 1.000

The contingency table below summarizes a survey of 1,000 bottled beverage consumers.

Find the following probabilities or percentages:
a) Probability that a consumer recycles beverage bottles. (Round your answer to 2 decimal places.)
b) Probability that a consumer who lives in a state with a deposit law does not recycle. (Round your answer to 2 decimal places.)
c) Percentage of consumers who recycle and live in a state with a deposit law. (Round your answer to 1 decimal place.)
d) Percentage of consumers in states with a deposit law who recycle. (Round your answer to the nearest whole percent.)

Explanation

(a)
P(Recycles) = 322/1,000 = 0.32
(b)
P(Don’t Recycle | Lives in Deposit Law State) = 67/200 = 0.34.
(c)
P(Recycles and Lives in Deposit Law State) = 133/1,000 = 0.133, or 13.3%.
(d)
P(Recycles | Lives in Deposit Law State) = 133/200 = 0.67, or 67%.

“The probability of rolling three sevens in a row with dice is .0046.”
Which kind of probability is it?

a) Empirical
b) Classical
c) Subjective

b) Classical

Which of the following could not be probability distributions?

a) None
b) Example A and C
c) Examples B and C
d) Example A and B

c) Examples B and C

Explanation

Example A is a probability distribution because the sum of P(x) is 1 (.8 + .2 = 1) and all probabilities are nonnegative.
Examples B and C are not probability distributions because the sum of P(x) is .95 for B and 1.30 for C.

Last year’s freshman class at Big State University totaled 5,331 students. Of those, 1,268 received a merit scholarship to help offset tuition costs their freshman year (although the amount varied per student). The amount a student received was N($3,446, $475). If the cost of full tuition was $4,300 last year, what percentage of students who received a merit scholarship did not receive enough to cover full tuition? (Round your answer to the nearest whole percent.)

Percentage of Students = 96%

Explanation

Find P(X < $4,300) = P(Z<4,300−3,446475)=P(Z<1.80)P(Z<4,300-3,446475)=P(Z<1.80) =NORM.S.DIST(1.80,1) = 0.9641. 96% of merit scholarship students will not receive enough to cover their full tuition.
Alternatively one could use the Excel function = NORM.DIST(4300,3446,475,1) = 0.9639. (The difference is due to the rounding of the z score to 1.80.)

Pepsi and Mountain Dew products sponsored a contest giving away a Lamborghini sports car worth $288,000. The probability of winning from a single bottle purchase was .00000510.

Find the expected value. (Round your answer to 4 decimal places.)

Expected Value = $1.4688

Explanation

X is the amount of the “value” of a bottle. X can be either $288,000, the value of the Lamborghini, or $0. To find the expected value multiply each value of X by the associated probability. E(X) = ($288,000)(.00000510) + ($0)(.99999490) = 1.4688.

Student Life Insurance Company wants to offer an insurance plan with a maximum claim amount of $6,700 for dorm students to cover theft of certain items. Past experience suggests that the probability of a maximum claim is .02. What premium should be charged if the company wants to make a profit of $36 per policy? Assume any student who files a claim files for the maximum amount and there is no deductible.

Premium Amount = $170

Explanation

X is the amount of the claim. Assume a student who files a claim will claim the maximum amount of $6,700. The values of X are either $6,700 or $0. Expected payout = E(X) = $6,700(.02) + ($0)(.98) = $134, so company adds $36 and charges $170.

The cabin of a business jet has a cabin height 5 feet 9 inches high.

If a business traveler’s height is N(5' 10", 2.7"), what percentage of the business travelers will have to stoop? (Round your answer to 4 decimal places.)

Probability = .6443

Explanation

We want to know the probability of getting a business traveler who is taller than 5’ 9”. To make it easier, we can convert everything to inches. Therefore, we should use a mean of 70 inches and a cabin height of 69 inches.

P(X>69)=1−P(X<69)P(X>69)=1−P(X<69) = 1 − NORM.DIST(69,70,2.7,1) = .6444. We can also use the standardized z-score:

z=69 −  702.7=−.37.  z=69⁢ −  702.7=−.37.⁢  

P(Z < −.37) = .3557. P(Z > −.37) = 1 − .3557 = .6443.

Use Excel to find each probability. (Round your answers to 4 decimal places.)

a) P(80 < X < 110) for N(100,15)
b) P(1.50 < X < 2.00) for N(0,1)
c) P(4,500 < X < 7,000) for N(6000,1000)
d) P(225 < X < 450) for N(600,100)

Explanation

a.
P(80 < X < 110) = P(X < 110) – P(X < 80) = NORM.DIST(110,100,15,1) − NORM.DIST(80,100,15,1) = .6563.
b.
P(1.50 < X < 2.00) = P(X < 2.00) – P(X < 1.50) = NORM.DIST(2,0,1,1) − NORM.DIST(1.5,0,1,1) = .0441.
c.
P(4500 < X < 7000) = P(X < 7000) – P(X < 4500) = NORM.DIST(7000,6000,1000,1) − NORM.DIST(4500,6000,1000,1) = .7745.
d.
P(225 < X < 450) = P(X < 450) – P(X < 225) = NORM.DIST(450,600,100,1) − NORM.DIST(225,600,100,1) = .0667.

The host of a daily financial radio program stated that in his opinion, “There is a 25 percent chance that AT&T Wireless and Verizon will merge.”

What kind of probability is this?

a) Empirical
b) Subjective
c) Classical

b) Subjective

On the midnight shift, the number of patients with head trauma in an emergency room has the probability distribution shown below.

a) Calculate the mean and standard deviation. (Round your mean value to 2 decimal places and standard deviation to 3 decimal places.)

b) Describe the shape of this distribution.
- Skewed to the right
- Skewed to the left
- Symmetric

Explanation

(a)
Based on the table below we see that: E(X) = 2.12, V(X) = 1.385600, σσ = 1.177.
x P(x)xP(x) x−E(X) P(x)[x−E(X)]² 0 .04 .00 −2.12 .179776 1 .33 .33 −1.12 .413952 2 .27 .54 −.12 .003888 3 .21 .63 .88 .162624 4 .13 .52 1.88 .459472 5 .02 .10 2.88 .165888 Total 1.00 2.12 1.385600
(b)
The distribution is skewed to the right.

Daily output of Marathon's Garyville, Lousiana, refinery is normally distributed with a mean of 232,000 barrels of crude oil per day with a standard deviation of 7,000 barrels.

a) What is the probability of producing at least 232,000 barrels? (Round your answer to 4 decimal places.)
b) What is the probability of producing between 232,000 and 239,000 barrels? (Round your answer to 4 decimal places.)
c) What is the probability of producing less than 239,000 barrels? (Round your answer to 4 decimal places.)
d) What is the probability of producing less than 245,000 barrels? (Round your answer to 4 decimal places.)
e) What is the probability of producing more than 225,000 barrels? (Round your answer to 4 decimal places.)

Explanation

(a)
P(X > 232,000) = 1 − P(X < 232,000) = 1 − NORM.DIST(232000,232000,7000,1) = 1− .50 = .50. We can also use the standardized z-score:

z =232,000 − 232,000 = 0.7,000
P(X < 232,000) = P(Z < 0) = .5 so P(X > 232,000) = 1 − .5 = .5.
(b)
P(X < 239,000) − P(X < 232,000) = NORM.DIST(239000,232000,7000,1) − NORM.DIST (232000,232000,7000,1) = .8413 − .5 = .3413. We can also use the standardized z-score:

z =239,000 − 232,000 = 1and z =232,000 − 232,000 = 0.7,0007,000P(232,000 < X < 239,000) = P(0 < Z < 1), which is .3413 (from Appendix C-1).
(c)
P(X < 239,000) = NORM.DIST(239000,232000,7000,1) = .8413. We can also use the standardized z-score:

z =239,000 − 232,000 = 1.7,000P(Z < 1) = .8413 (from Appendix C-2).
(d)
P(X < 245,000) = NORM.DIST(245000,232000,7000,1) = .9684. We can also use the standardized z-score:

z =245,000 − 232,000 = 1.86.7,000P(Z < 1.86) = .9686 (from Appendix C-2).
(e)
P(X > 225,000) = 1 − P(X < 225,000) = 1 − NORM.DIST(225000,232000,7000,1) = 1 − .1587 = .8413. We can also use the standardized z-score:

z =225,000 − 232,000 = −1.7,000
1 − P(Z < −1) = .8413.

The following contingency table shows average yield (rows) and average duration (columns) for 38 bond funds.

For a randomly chosen bond fund, find the probability of the following: (Round your answers to 4 decimal places.)

a) The bond fund is of long duration.
b) The bond fund has high yield.
c) The bond fund has high yield given that it is of short duration.
d) The bond fund is of short duration given that it has high yield.

Explanation

a. P(D ₃) = 19/38 = 0.5000.
b. P (Y ₃) = 21/38 = 0.5526.
c. P(Y ₃ | D ₁) = P(Y ₃ ∩∩ D ₁)/(D ₁) = (4/38)/(11/38) = 0.3636. Another way is 4/11 = 0.3636 because the two 38s cancel each other out.
d. P(D ₁ | Y ₃) = P(Y ₃ ∩∩ D ₁)/P(Y ₃) = (4/38)/(21/38) = 0.1905. Another way is 4/21 = 0.1905 because the two 38s cancel each other out.

If Punxsutawney Phil sees his shadow on February 2, then legend says that winter will last 6 more weeks. In 123 years, Phil has seen his shadow 103 times.

a) What is the probability that Phil will see his shadow on a randomly chosen Groundhog Day? (Round your answer to 4 decimal places.)

b) What kind of probability is this?
- Empirical
- Classical
- Subjective

Explanation

a)
103/123 = 0.8374.

b)
Empirical

The contingency table below shows the results of a survey of video viewing habits by age.

a) Probability that a viewer is aged 18–34. (Round your answer to 2 decimal places.)
b) Probability that a viewer prefers watching videos on a TV screen. (Round your answer to 2 decimal places.)
c) Percentage of viewers who are 18–34 and prefer videos on a mobile or laptop device.
d) Percentage of viewers given they are 18–34 who prefer videos on a mobile or laptop device. (Round your answer to 2 decimal places.)
e) Percentage of viewers who are 35–54 or prefer videos on a mobile or laptop device.

Explanation

(a)
P(A viewer is aged 18−34) = 60/100 = 0.60.
(b)
P(A viewer prefers watching TV videos) = 50/100 = 0.50.
(c)
P (A viewer is aged 18−34 and prefers watching videos on mobile/laptop device) = 32/100 = 0.32 or 32%.
(d)
In Image
(e)
P(Aged 35−54 ∪∪ mobile/laptop device) = P(Aged 35−54) + P(mobile/laptop device ) – P(Aged 35−54 ∩∩ mobile/laptop device) = (0.28 + 0.50) – 0.14 = 0.64 or 64%

Suppose X ~ N(58, 7). Write the Excel function to calculate the following probabilities:

a) P(X<55.1)
b) P(X ≥ 60.6)
c) P(54.1 ≤ X < 63.4)

a) =NORM.DIST(55.1,58,7,1)
b) =1−NORM.DIST(60.6,58,7,1)
c) =NORM.DIST(63.4,58,7,1) – NORM.DIST(54.1,58,7,1)

Which of the following could not be probability distributions?

a) Example B and C
b) Example A and B
c) Example A and C
d) None

c) Example A and C

Explanation

Example B is a probability distribution because the sum of P(x) is 1 (.1 + .2 + .4 + .2 + .1 1) and all probabilities are nonnegative.
Example A and C are not probability distributions because the sum of P(x) is 1.10 for A and 1.40 for C.

Oxnard Petro Ltd. is buying hurricane insurance for its off-coast oil drilling platform. During the next five years, the probability of total loss of only the above-water superstructure ($220 million) is .30, the probability of total loss of the facility ($920 million) is .30, and the probability of no loss is .40.

Find the expected loss. (Input the amount as a positive value.)

Expected Loss = $342 million

Explanation

Let X equal the loss during a hurricane. The values of X are either $220 million, $920 million, or $0. Expected Loss = E(X ) = $220(.30) + $920(.30) + $0(.40) = $342 million.

On hot, sunny, summer days, Jane rents inner tubes by the river that runs through her town. Based on her past experience, she has assigned the following probability distribution to the number of tubes she will rent on a randomly selected day.

a) Calculate the expected value and standard deviation of this random variable X.
b) Describe the shape of this distribution.

Explanation

(a)
Based on the table below we see that: E(X) = 58.25, V(X) = 513.188, σσ = 22.65.
(b)
The distribution is skewed to the right.

Researchers examined forecasters’ interest rate predictions for 34 quarters to see whether the predictions corresponded to what actually happened. The 2 × 2 contingency table below shows the frequencies of actual and predicted interest rate movements.

Calculate each probability (i–vi): (Round your answers to 4 decimal places.)

Explanation
i. P(F−) = 20/38 = 0.5263. The probability that the forecasters predicted a decline in interest rates is 0.5263.

ii. P(A+) = 21/38 = 0.5526. The probability there was a rise in interest rates is 0.5526.

iii. P(A− | F−) = 8/20 = 0.4000. Given that the forecasters predicted a decline in interest rates, the probability that there was an actual decline is 0.4000.

iv. P(A+ | F+) = 9/18 = 0.5000. Given that the forecasters predicted an increase in interest rates, the probability that there was an actual increase is 0.5000.

v. P(A+ and F+) = 9/38 = 0.2368. The probability that in a given year there was both a forecasted increase and actual increase in interest rates is 0.2368.

vi. P(A− and F−) = 8/38 = 0.2105. The probability that in a given year there was both a forecasted decline and actual decline in interest rates is 0.2105.

If the weight (in grams) of cereal in a box of Lucky Charms is N(487,7), what is the probability that the box will contain less than the advertised weight of 457 g? Note: You may need to use Excel to calculate the exact probabilities. (Round your answer to 5 decimal places.)

Probability = .00001

Explanation

P(X < 457)  =  P(Z <  457 −  4877)  =  P(Z<−4.29)  =  .00001.P(X⁢ < 457)  =  P(Z⁢ <  457⁢ −  4877)  =  P(Z<-4.29)  =  .00001.

Using Excel: NORM.DIST(457,487,7,1) = .00001.

It is very unlikely that a box will contain less than 457 grams.