Chapter Notes

Key Terms

Note: Starred (*) terms are from the optional sections in this chapter.

Key Symbols

p(x) Probability distribution for discrete random variable x
f(x) Probability distribution for continuous random variable x
S Outcome of binomial trial denoted “success”
F Outcome of binomial trial denoted “failure”
p P(S) in binomial trial
q P(F) inbinomialtrial=1p
f(x) Probability distribution (density function) for continuous random variable x
e Constant used in normal probability distributions, e=2.71828
π Constant used in normal probability distributions, π=3.1416
μx True mean of sampling distribution of x
σx True standard deviation of sampling distribution of x

Key Ideas

Properties of Discrete Probability Distributions

  1. p(x)0

  2. all xp(x)=1

Properties of Continuous Probability Distributions

  1. P(x=a)=0

  2. P(a<x<b) is area under curve between a and b

Methods for Assessing Normality

  1. (1) Histogram

  2. (2) Stem-and-leaf display

  3. (3) (IQR)/1.3

  4. (4) Normal probability plot

Normal Approximation to Binomial

x is binomial (n, p)

P(a)P{z<(a+.5)μ}
  • Sampling distribution of a statistic—the theoretical probability distribution of the statistic in repeated sampling

  • Central Limit Theorem—the sampling distribution of the sample mean, x, or the sample proportion, p^, is approximately normal for large n

Key Formulas

Alternate View
Random Variable Probability Distribution Mean Variance z-Score
General Discrete: Table, formula, or graph for p(x) all xxp(x) all x(xμ)2p(x)
Binomial: p(x)=(nx)pxqnx np npq
Normal: f(x)=1σ2πe1/2[(xμ)/σ]2 μ σ2
Standard Normal: f(z)=12πe1/2(z)2z=(xμ)/σ μ=0 σ2=1
Sampling distribution of x μx=μ σx¯=σn z=x¯μx¯σx¯=x¯μσ/n

Guide to Selecting a Probability Distribution

Generating the Sampling Distribution of x¯

Supplementary Exercises 4.176–4.220

Note: Starred (*) exercises refer to the optional sections in this chapter.

Understanding the Principles

  1. 4.176 Which of the following describe discrete random variables, and which describe continuous random variables?

    1. The length of time that an exercise physiologist’s program takes to elevate her client’s heart rate to 140 beats per minute

    2. The number of crimes committed on a college campus per year

    3. The number of square feet of vacant office space in a large city

    4. The number of voters who favor a new tax proposal

  2. 4.177 For each of the following examples, decide whether x is a binomial random variable and explain your decision:

    1. A manufacturer of computer chips randomly selects 100 chips from each hour’s production in order to estimate the proportion of defectives. Let x represent the number of defectives in the 100 chips sampled.

    2. Of five applicants for a job, two will be selected. Although all applicants appear to be equally qualified, only three have the ability to fulfill the expectations of the company. Suppose that the two selections are made at random from the five applicants, and let x be the number of qualified applicants selected.

    3. A software developer establishes a support hot line for customers to call in with questions regarding use of the software. Let x represent the number of calls received on the hot line during a specified workday.

    4. Florida is one of a minority of states with no state income tax. A poll of 1,000 registered voters is conducted to determine how many would favor a state income tax in light of the state’s current fiscal condition. Let x be the number in the sample who would favor the tax.

  3. 4.178 Describe how you could obtain the simulated sampling distribution of a sample statistic.

  4. 4.179 True or False. The sample mean, x, will always be equal to μx.

  5. 4.180 True or False. The sampling distribution of x is normally distributed, regardless of the size of the sample n. 

  6. 4.181 True or False. The standard error of x will always be smaller than σ.

  7. 4.182 Consider the discrete probability distribution shown here:

    Alternate View
    x 10 12 18 20
    p(x) .2 .3 .1 .4
    1. Calculate μ,σ2, and σ.

    2. What is P(x<15)?

    3. Calculate μ±2σ.

    4. What is the probability that x is in the interval μ±2σ?

  8. 4.183 Suppose x is a binomial random variable with n=20 and p=.7.

    1. Find P(x=14).

    2. Find P(x12).

    3. Find P(x>12).

    4. Find P(9x18).

    5. Find P(8<x<18).

    6. Find μ,σ2, and σ.

    7. What is the probability that x is in the interval μ±2σ?

  9. 4.184 The random variable x has a normal distribution with μ=70 and σ=10. Find the following probabilities:

    1. P(x75)

    2. P(x90)

    3. P(60x75)

    4. P(x>75)

    5. P(x=75)

    6. P(x95)

  10. 4.185 A random sample of n=68 observations is selected from a population with μ=19.6 and σ=3.2. Approximate each of the following probabilities:

    1. P(x19.6)

    2. P(x19)

    3. P(x20.1)

    4. P(19.2x20.6)

  11. *4.186 Assume that x is a binomial random variable with n=100 and p=.5. Use the normal probability distribution to approximate the following probabilities:

    1. P(x48)

    2. P(50x65)

    3. P(x70)

    4. P(55x58)

    5. P(x=62)

    6. P(x49 or x72)

Apply the Concepts–Basic

  1. 4.187 Use of laughing gas. According to the American Dental Association, 60% of all dentists use nitrous oxide (“laughing gas”) in their practice. If x equals the number of dentists in a random sample of five dentists who use laughing gas in practice, then the probability distribution of x is

    Alternate View
    x 0 1 2 3 4 5
    p(x) .0102 .0768 .2304 .3456 .2592 .0778
    1. Find the probability that the number of dentists using laughing gas in the sample of five is less than 2.

    2. Find E(x) and interpret the result.

    3. Show that the distribution of x is binomial with n=5 and p=.6.

  2. 4.188 Parents who condone spanking. According to a nationwide survey, 60% of parents with young children condone spanking their child as a regular form of punishment (Tampa Tribune, Oct. 5, 2000). Consider a random sample of three people, each of whom is a parent with young children. Assume that x, the number in the sample who condone spanking, is a binomial random variable.

    1. What is the probability that none of the three parents condones spanking as a regular form of punishment for their children?

    2. What is the probability that at least one condones spanking as a regular form of punishment?

    3. Give the mean and standard deviation of x. Interpret the results.

    4. A child psychologist with 150 parent clients claims that no more than 20 of the parents condone spanking. Do you believe this claim? Explain.

  3. 4.189 Tracking missiles with satellite imagery. The Space-Based Infrared System (SBIRS) uses satellite imagery to detect and track missiles (Chance, Summer 2005). The probability that an intruding object (e.g., a missile) will be detected on a flight track by SBIRS is .8. Consider a sample of 20 simulated tracks, each with an intruding object. Let x equal the number of these tracks on which SBIRS detects the object.

    1. Demonstrate that x is (approximately) a binomial random variable.

    2. Give the values of p and n for the binomial distribution.

    3. Find P(x=15), the probability that SBIRS will detect the object on exactly 15 tracks.

    4. Find P(x15), the probability that SBIRS will detect the object on at least 15 tracks.

    5. Find E(x) and interpret the result.

  4. 4.190 Married-women study. According to Women’s Day magazine (Jan. 2007), only 50% of married women would marry their current husbands again if given the chance. Consider the number x in a random sample of 10 married women who would marry their husbands again. Identify the discrete probability distribution that best models the distribution of x. Explain.

  5. 4.191 Dust mite allergies. A dust mite allergen level that exceeds 2 micrograms per gram (μg/g) of dust has been associated with the development of allergies. Consider a random sample of four homes, and let x be the number of homes with a dust mite level that exceeds 2μg/g. The probability distribution for x, based on a study by the National Institute of Environmental Health Sciences, is shown in the following table:

    Alternate View
    x   0   1   2   3   4
    p(x) .09 .30 .37 .20 .04
    1. Verify that the probabilities for x in the table sum to 1.

    2. Find the probability that three or four of the homes in the sample have a dust mite level that exceeds 2μg/g.

    3. Find the probability that fewer than two homes in the sample have a dust mite level that exceeds 2μg/g.

    4. Find E(x). Give a meaningful interpretation of the result.

    5. Find σ.

    6. Find the exact probability that x is in the interval μ±2σ. Compare your answer with Chebyshev’s rule and the empirical rule.

Applying the Concepts—Basic

  1. CRASH 4.192 NHTSA crash safety tests. Refer toConsider the National Highway Traffic Safety Administration (NHTSA) crash test data for new cars, presented in Exercise 2.190 (p. 107). One of the variables measured is the severity of a driver’s head injury when the car is in a head-on collision with a fixed barrier while traveling at 35 miles per hour. The more points assigned to the head injury rating, the more severe is the injury. The head injury ratings can be shown to be approximately normally distributed with a mean of 605 points and a standard deviation of 185 points. One of the crash-tested cars is randomly selected from the data, and the driver’s head injury rating is observed.

    1. Find the probability that the rating will fall between 500 and 700 points.

    2. Find the probability that the rating will fall between 400 and 500 points.

    3. Find the probability that the rating will be less than 850 points.

    4. Find the probability that the rating will exceed 1,000 points.

    5. What rating will only 10% of the crash-tested cars exceed?

  2. 4.193 Belief in an afterlife. A national poll conducted by The New York Times (May 7, 2000) revealed that 80% of Americans believe that after you die, some part of you lives on, either in a next life on earth or in heaven. Consider a random sample of 10 Americans and count x, the number who believe in life after death.

    1. Find P(x=3).

    2. Find P(x7).

    3. Find P(x>4).

  3. 4.194 Improving SAT scores. Refer toConsider the Chance (Winter 2001) study of students who paid a private tutor to help them improve their SAT scores, presented in Exercise 2.197 (p. 108). The table summarizing the changes in both the SAT-Mathematics and SAT-Verbal scores for these students is reproduced here. Assume that both distributions of SAT score changes are approximately normal.

    SAT-Math SAT-Verbal
    Mean change in score 19  7
    Standard deviation of changes in score 65 49
    1. What is the probability that a student increases his or her score on the SAT-Math test by at least 50 points?

    2. What is the probability that a student increases his or her score on the SAT-Verbal test by at least 50 points?

  4. 4.195 Passing the FCAT math test. All Florida high schools require their students to demonstrate competence in mathematics by scoring 70% or above on the FCAT mathematics achievement test. The FCAT math scores of those students taking the test for the first time are normally distributed with a mean of 77% and a standard deviation of 7.3%. What percentage of students who take the test for the first time will pass it? 

  5. *4.196 Where will you get your next pet? Refer to Exercise 5.86 (p. 266), Recal that 50% of pet owners will get their next dog or cat from a shelter (USA Today, May 12, 2010). Now, consider a random sample of 200 pet owners, and let x be the number who will get their next dog or cat from a shelter.

    1. Find the mean of x.

    2. Find the standard deviation of x.

    3. Find the z-score for the value x=110.

    4. Find the approximate probability that x is less than or equal to 110.

  6. 4.197 Transmission delays in wireless technology. In Mobile Networks and Applications (Dec. 2003), Resource Reservation Protocol (RSVP) was applied to mobile wireless technology (e.g., a PC notebook with wireless LAN card for Internet access). A simulation study revealed that the transmission delay (measured in milliseconds) of an RSVP–linked wireless device has an approximate normal distribution with mean μ=48.5 milliseconds and σ=8.5 milliseconds.

    1. What is the probability that the transmission delay is less than 57 milliseconds?

    2. What is the probability that the transmission delay is between 40 and 60 milliseconds?

  7. 4.198 Salaries of travel managers. According to a 2012 Business Travel News survey, the average salary of a travel manager is $110,550. Assume that the standard deviation of such salaries is $30,000. Consider a random sample of 50 travel managers, and let x represent the mean salary for the sample.

    1. What is μx?

    2. What is σx?

    3. Describe the shape of the sampling distribution of x.

    4. Find the z-score for the value x¯=$100,000.

    5. Find P(x>100,000).

  8. 4.199 Children’s attitudes toward reading. In the journal Knowledge Quest (Jan./Feb. 2002), education professors investigated children’s attitudes toward reading. One study measured third through sixth graders’ attitudes toward recreational reading on a 140-point scale (where higher scores indicate a more positive attitude). The mean score for this population of children was 106 points, with a standard deviation of 16.4 points. Consider a random sample of 36 children from this population, and let x represent the mean recreational reading attitude score for the sample.

    1. What is μx?

    2. What is σx?

    3. Describe the shape of the sampling distribution of x.

    4. Find the z-score for the value x¯=$100,000

    5. Find P(x<100).

  9. 4.200 Research on eating disorders. Refer to The American Statistician (May 2001) study of female students who suffer from bulimia, presented in Exercise 2.44 (p. 52). Recall that each student completed a questionnaire from which a “fear of negative evaluation” (FNE) score was produced. (The higher the score, the greater is the fear of negative evaluation.) Suppose the FNE scores of bulimic students have a distribution with mean μ=18 and standard deviation σ=5. Now, consider a random sample of 45 female students with bulimia.

    1. What is the probability that the sample mean FNE score is greater than 17.5?

    2. What is the probability that the sample mean FNE score is between 18 and 18.5?

    3. What is the probability that the sample mean FNE score is less than 18.5?

Applying the Concepts—Intermediate

  1. 4.201 When to replace a maintenance system. An article in the Journal of Quality of Maintenance Engineering (Vol. 19, 2013) studied the problem of finding the optimal replacement policy for a maintenance system. Consider a system that is tested every 12 hours. The test will determine whether there are any flaws in the system. Assume the probability of no flaw being detected is .85. If a flaw (failure) is detected, the system is repaired. Following the 5th failed test, the system is completely replaced. Now let x represent the number of tests until the system needs to be replaced.

    1. Give the probability distribution for x as a formula.

    2. Find the probability that the system needs to be replaced after 8 total tests.

  2. 4.202 Student gambling on sports. A study of gambling activity at the University of West Georgia (UWG) discovered that 60% of the male students wagered on sports the previous year (The Sport Journal, Fall 2006). Consider a random sample of 50 UWG male students. How many of these students would you expect to have gambled on sports the previous year? Give a range that is likely to include the number of male students who have gambled on sports. 

  3. 4.203 Parents’ behavior at a gym meet. Pediatric Exercise Science (Feb. 1995) published an article on the behavior of parents at competitive youth gymnastic meets. On the basis of a survey of the gymnasts, the researchers estimated the probability of a parent “yelling” at his or her child before, during, or after the meet as .05. In a random sample of 20 parents attending a gymnastic meet, find the probability that at least 1 parent yells at his or her child before, during, or after the meet. 

  4. *4.204 Countries that allow a free press. The degree to which democratic and nondemocratic countries attempt to control the news media was examined in the Journal of Peace Research (Nov. 1997). The article reported that 80% of all democratic regimes allow a free press. In contrast, 10% of all nondemocratic regimes allow a free press.

    1. In a random sample of 50 democratic regimes, how many would you expect to allow a free press? Give a range that is highly likely to include the number of democratic regimes with a free press.

    2. In a random sample of 50 nondemocratic regimes, how many would you expect to allow a free press? Give a range that is highly likely to include the number of nondemocratic regimes with a free press.

  5. 4.205 Chickens with fecal contamination. The United States Department of Agriculture (USDA) reports that, under its standard inspection system, one in every 100 slaughtered chickens passes inspection for fecal contamination (Tampa Tribune, Mar. 31, 2000). In Exercise 3.21 (p. 128), you found the probability that a randomly selected slaughtered chicken passes inspection for fecal contamination. Now find the probability that, in a random sample of 5 slaughtered chickens, at least one passes inspection for fecal contamination.

  6. 4.206 Comparison of exam scores: red versus blue exam. Refer to the Teaching Psychology (May 1998) study of how external clues influence performance, presented in Exercise 2.128 (p. 82). Recall that two different forms of a midterm psychology examination were given, one printed on blue paper and the other on red paper. Grading only the difficult questions, the researchers found that scores on the blue exam had a distribution with a mean of 53% and a standard deviation of 15%, while scores on the red exam had a distribution with a mean of 39% and a standard deviation of 12%. Assuming that both distributions are approximately normal, on which exam is a student more likely to score below 20% on the difficult questions, the blue one or the red one? (Compare your answer with that of Exercise 2.128 c.) 

  7. 4.207 Visually impaired students. The Journal of Visual Impairment & Blindness (May–June 1997) published a study of the lifestyles of visually impaired students. Using diaries, the students kept track of several variables, including number of hours of sleep obtained in a typical day. These visually impaired students had a mean of 9.06 hours and a standard deviation of 2.11 hours. Assume that the distribution of the number of hours of sleep for this group of students is approximately normal.

    1. Find the probability that a visually impaired student obtains less than 6 hours of sleep on a typical day.

    2. Find the probability that a visually impaired student gets between 8 and 10 hours of sleep on a typical day.

    3. Twenty percent of all visually impaired students obtain less than how many hours of sleep on a typical day?

  8. 4.208 Length of gestation for pregnant women. On the basis of data from the National Center for Health Statistics, N. Wetzel used the normal distribution to model the length of gestation for pregnant U.S. women (Chance, Spring 2001). Gestation has a mean length of 280 days with a standard deviation of 20 days.

    1. Find the probability that the length of gestation is between 275.5 and 276.5 days. (This estimate is the probability that a woman has her baby 4 days earlier than the “average” due date.)

    2. Find the probability that the length of gestation is between 258.5 and 259.5 days. (This estimate is the probability that a woman has her baby 21 days earlier than the “average” due date.)

    3. Find the probability that the length of gestation is between 254.5 and 255.5 days. (This estimate is the probability that a woman has her baby 25 days earlier than the “average” due date.)

    4. The Chance article referenced a newspaper story about three sisters who all gave birth on the same day (Mar. 11, 1998). Karralee had her baby 4 days early, Marrianne had her baby 21 days early, and Jennifer had her baby 25 days early. Use the results from parts ac to estimate the probability that three women have their babies 4, 21, and 25 days early, respectively. Assume that the births are independent events.

  9. PANEL 4.209 Wear-out of used display panels. Wear-out failure time of electronic components is often assumed to have a normal distribution. Can the normal distribution be applied to the wear-out of used manufactured products, such as colored display panels? A lot of 50 used display panels was purchased by an outlet store. Each panel displays 12 to 18 color characters. Prior to acquisition, the panels had been used for about one-third of their expected lifetimes. The data in the table below give the failure times (in years) of the 50 used panels. Use the techniques of this chapter to determine whether the used panel wear-out times are approximately normally distributed. 

  10. 4.210 Fitness of cardiac patients. The physical fitness of a patient is often measured by the patient’s maximum oxygen uptake (recorded in milliliters per kilogram, ml/kg). The mean maximum oxygen uptake for cardiac patients who regularly participate in sports or exercise programs was found to be 24.1, with a standard deviation of 6.30 (Adapted Physical Activity Quarterly, Oct. 1997). Assume that this distribution is approximately normal.

    1. What is the probability that a cardiac patient who regularly participates in sports has a maximum oxygen uptake of at least 20 ml/kg?

    2. What is the probability that a cardiac patient who regularly exercises has a maximum oxygen uptake of 10.5 ml/kg or lower?

    3. Consider a cardiac patient with a maximum oxygen uptake of 10.5. Is it likely that this patient participates regularly in sports or exercise programs? Explain.

  11. 4.211 Susceptibility to hypnosis. The Computer-Assisted Hypnosis Scale (CAHS) is designed to measure a person’s susceptibility to hypnosis. CAHS scores range from 0 (no susceptibility) to 12 (extremely high susceptibility). A study in Psychological Assessment (Mar. 1995) reported a mean CAHS score of 4.59 and a standard deviation of 2.95 for University of Tennessee undergraduates. Assume that μ=4.59 and σ=2.95 for this population. Suppose a psychologist uses the CAHS to test a random sample of 50 subjects.

    Data for Exercise 4.209

    Alternate View
    0.01 1.21 1.71 2.30 2.96 0.19 1.22 1.75 2.30 2.98 0.51
    1.24 1.77 2.41 3.19 0.57 1.48 1.79 2.44 3.25 0.70 1.54
    1.88 2.57 3.31 0.73 1.59 1.90 2.61 1.19 0.75 1.61 1.93
    2.62 3.50 0.75 1.61 2.01 2.72 3.50 1.11 1.62 2.16 2.76
    3.50 1.16 1.62 2.18 2.84 3.50

    Based on Irony, T. Z., Lauretto, M., Pereira, C., and Stern, J. M. “A Weibull wearout test: Full Bayesian approach,” paper presented at Mathematical Sciences Colloquium, Binghamton University, Binghamton, UK, Dec. 2001.

    1. Would you expect to observe a sample mean CAHS score of x=6 or higher? Explain.

    2. Suppose the psychologist actually observes x=6.2. On the basis of your answer to part a, make an inference about the population from which the sample was selected.

  12. 4.212 Supercooling temperature of frogs. Many species of terrestrial tree frogs can survive prolonged exposure to low winter temperatures during hibernation. In freezing conditions, the frog’s body temperature is called its supercooling temperature. A study in Science revealed that the supercooling temperature of terrestrial frogs frozen at 6°C has a relative frequency distribution with a mean of 2°C and a standard deviation of .3°C. Consider the mean supercooling temperature x of a random sample of n=42 terrestrial frogs frozen at 6°C.

    1. Find the probability that x exceeds 2.05°C.

    2. Find the probability that x falls between 2.20°C and 2.10°C.

  13. 4.213 Levelness of concrete slabs. Geotechnical engineers use water-level “manometer” surveys to assess the levelness of newly constructed concrete slabs. Elevations are typically measured at eight points on the slab; of interest is the maximum differential between elevations. The Journal of Performance of Constructed Facilities (Feb. 2005) published an article on the levelness of slabs in California residential developments. Elevation data collected on over 1,300 concrete slabs before tensioning revealed that the maximum differential x has a mean of μ=.53 inch and a standard deviation of σ=.193inch. Consider a sample of n=50 slabs selected from those surveyed, and let x represent the mean of the sample.

    1. Describe fully the sampling distribution of x.

    2. Find P(x>.58).

    3. The study also revealed that the mean maximum differential of concrete slabs measured after tensioning and loading is μ=.58 inch. Suppose the sample data yield x=.59 inch. Comment on whether the sample measurements were obtained before tensioning or after tensioning and loading.

Applying the Concepts—Advanced

  1. 4.214 Network forensic analysis. A network forensic analyst is responsible for identifying worms, viruses, and infected nodes in the computer network. A new methodology for finding patterns in data that signify infections was investigated in IEEE Transactions on Information Forensics and Security (May 2013). The method uses multiple filters to check strings of information. For this exercise, consider a data string of length 4 bytes (positions), where each byte is either a 0 or a 1 (e.g., 0010). Also, consider two possible strings, named S1 and S2. In a simple single filter system, the probability that S1 and S2 differ in any one of the bytes is .5. Derive a formula for the probability that the two strings differ on exactly x of the 4 bytes. Do you recognize this probability distribution?

  2. 4.215 How many questionnaires to mail? The probability that a person responds to a mailed questionnaire is .4. How many questionnaires should be mailed if you want to be reasonably certain that at least 100 will be returned?

  3. 4.216 Dye discharged in paint. A machine used to regulate the amount of dye dispensed for mixing shades of paint can be set so that it discharges an average of μ milliliters (mL) of dye per can of paint. The amount of dye discharged is known to have a normal distribution with a standard deviation of .4mL. If more than 6mL of dye are discharged when making a certain shade of blue paint, the shade is unacceptable. Determine the setting for μ so that only 1% of the cans of paint will be unacceptable. 

  4. 4.217 Soft-drink bottles. A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have an internal pressure of at least 150 pounds per square inch (psi). A prospective bottle vendor claims that its production process yields bottles with a mean internal pressure of 157 psi and a standard deviation of 3 psi. The bottler strikes an agreement with the vendor that permits the bottler to sample from the vendor’s production process to verify the vendor’s claim. The bottler randomly selects 40 bottles from the last 10,000 produced, measures the internal pressure of each, and finds the mean pressure for the sample to be 1.3 psi below the process mean cited by the vendor.

    1. Assuming the vendor’s claim to be true, what is the probability of obtaining a sample mean this far or farther below the process mean? What does your answer suggest about the validity of the vendor’s claim?

    2. If the process standard deviation were 3 psi as claimed by the vendor, but the mean were 156 psi, would the observed sample result be more or less likely than the result in part a? What if the mean were 158 psi?

    3. If the process mean were 157 psi as claimed, but the process standard deviation were 2 psi, would the sample result be more or less likely than the result in part a? What if instead the standard deviation were 6 psi?

Critical Thinking Challenge

  1. *4.218 Space shuttle disaster. On January 28, 1986, the space shuttle Challenger exploded, killing all seven astronauts aboard. An investigation concluded that the explosion was caused by the failure of the O ring seal in the joint between the two lower segments of the right solid rocket booster. In a report made one year prior to the catastrophe, the National Aeronautics and Space Administration (NASA) claimed that the probability of such a failure was about 160,000, or about once in every 60,000 flights. But a risk-assessment study conducted for the Air Force at about the same time assessed the probability to be 135, or about once in every 35 missions. (Note: The shuttle had flown 24 successful missions prior to the disaster.) Given the events of January 28, 1986, which risk assessment—NASA’s or the Air Force’s—appears to be more appropriate?

  2. 4.219 IQs and The Bell Curve. The controversial book, The Bell Curve (Free Press, 1994), employs statistical analyses heavily in an attempt to support the authors’ position on “intelligence and class structure in American life.” Since the book’s publication, many expert statisticians have raised doubts about the authors’ statistical methods and the inferences drawn from them. (See, for example, “Wringing The Bell Curve: A cautionary tale about the relationships among race, genes, and IQ,” Chance, Summer 1995.) One of the many controversies sparked by the book is the authors’ tenet that level of intelligence (or lack thereof) is a cause of a wide range of intractable social problems, including constrained economic mobility. The measure of intelligence chosen by the authors is the well-known intelligence quotient (IQ). Psychologists traditionally treat IQ as a random variable having a normal distribution with mean μ=100 and standard deviation σ=15.

    The Bell Curve refers to five cognitive classes of people defined by percentiles of the normal distribution. Class I (“very bright”) consists of those with IQs above the 95th percentile; Class II (“bright”) are those with IQs between the 75th and 95th percentiles; Class III (“normal”) includes IQs between the 25th and 75th percentiles; Class IV (“dull”) are those with IQs between the 5th and 25th percentiles; and Class V (“very dull”) are IQs below the 5th percentile.

    1. Assuming that the distribution of IQ is accurately represented by the normal curve, determine the proportion of people with IQs in each of the five cognitive classes.

    2. Although the authors define the cognitive classes in terms of percentiles, they stress that IQ scores should be compared with z-scores, not percentiles. In other words, it is more informative to give the difference in z-scores for two IQ scores than it is to give the difference in percentiles. Do you agree?

    3. Researchers have found that scores on many intelligence tests are decidedly nonnormal. Some distributions are skewed toward higher scores, others toward lower scores. How would the proportions in the five cognitive classes defined in part a differ for an IQ distribution that is skewed right? Skewed left?

  3. 4.220 Fecal pollution at Huntington Beach. The state of California mandates fecal indicator bacteria monitoring at all public beaches. When the concentration of fecal bacteria in the water exceeds a certain limit (400 colony-forming units of fecal coliform per 100 milliliters), local health officials must post a sign (called surf zone posting) warning beachgoers of potential health risks upon entering the water. For fecal bacteria, the state uses a single-sample standard; that is, if the fecal limit is exceeded in a single sample of water, surf zone posting is mandatory. This single-sample standard policy has led to a recent rash of beach closures in California.

    Joon Ha Kim and Stanley B. Grant, engineers at the University of California at Irvine, conducted a study of the surf water quality at Huntington Beach in California and reported the results in Environmental Science & Technology (Sept. 2004). The researchers found that beach closings were occurring despite low pollution levels in some instances, while in others, signs were not posted when the fecal limit was exceeded. They attributed these “surf zone posting errors” to the variable nature of water quality in the surf zone (for example, fecal bacteria concentration tends to be higher during ebb tide and at night) and the inherent time delay between when a water sample is collected and when a sign is posted or removed. In order to prevent posting errors, the researchers recommend using an averaging method, rather than a single sample, to determine unsafe water quality. (For example, one simple averaging method is to take a random sample of multiple water specimens and compare the average fecal bacteria level of the sample with the limit of 400cfu/100mL in order to determine whether the water is safe.)

    Discuss the pros and cons of using the single-­sample standard versus the averaging method. Part of your discussion should address the probability of posting a sign when in fact the water is safe and the probability of posting a sign when in fact the water is unsafe. (Assume that the fecal bacteria concentrations of water specimens at Huntington Beach follow an approximately normal distribution.)

References

  • Hogg, R. V., McKean, J. W., and Craig, A. T. Introduction to Mathematical Statistics, 6th ed. Upper Saddle River, NJ: Prentice Hall, 2005.

  • Larsen, R. J., and Marx, M. L. An Introduction to Mathematical Statistics and Its Applications, 4th ed. Upper Saddle River, NJ: Prentice Hall, 2005.

  • Lindgren, B. W. Statistical Theory, 3rd ed. New York: Macmillan, 1976.

  • Lindgren, B. W. Statistical Theory, 4th ed. New York: Chapman & Hall, 1993.

  • Mood, A. M., Graybill, F. A., and Boes, D. C. Introduction to the Theory of Statistics, 3rd ed. New York: McGraw-Hill, 1974.

  • Parzen, E. Modern Probability Theory and Its Applications. New York: Wiley, 1960.

  • Ramsey, P. P., and Ramsey, P. H. “Simple tests of normality in small samples.” Journal of Quality Technology, Vol. 22, 1990.

  • Ross, S. M. Stochastic Processes, 2nd ed. New York: Wiley, 1996.

  • Wackerly, D., Mendenhall, W., and Scheaffer, R. L. Mathematical Statistics with Applications, 7th ed. North Scituate, MA: Duxbury, 2008.

  • Winkler, R. L., and Hays, W. Statistics: Probability, Inference, and Decision, 2nd ed. New York: Holt, Rinehart and Winston, 1975.

Using Technology MINITAB: Binomial Probabilities, Normal Probability, and Simulated Sampling Distribution

Binomial Probabilites

  1. Step 1 Select the “Calc” button on the MINITAB menu bar, click on “Probability Distributions,” and then finally select the discrete distribution of your choice (e.g., “Binomial”), as shown in Figure 4.M.1.

    Figure 4.M.1

    MINITAB menu options for discrete probabilities

  2. Step 2 Select either “Probability” or “Cumulative probability” on the resulting dialog box.

  3. Step 3 Specify the parameters of the distribution (e.g., sample size n, probability of success p).

    [Note: For the Poisson option, enter the mean λ. For the hypergeometric option, enter N, r, and n.]

  4. Step 4 Specify the value x of interest in the “Input constant” box.

  5. Step 5 Click “OK.” The probability for the value of x will appear on the MINITAB session window.

    [Note: Figure 4.M.2 gives the specifications for finding p(x=2) in a binomial distribution with n=5 and p=.3.]

    Figure 4.M.2

    MINITAB binomial distribution dialog box

Normal Probabilities

  1. Step 1 Select the “Calc” button on the MINITAB menu bar, click on “Probability Distributions,” and then finally select the Continuous distribution of your choice (e.g., “Normal”), as shown in Figure 4.M.3.

  2. Step 2 Select either “Probability” or “Cumulative probability” on the resulting dialog box.

  3. Step 3 Specify the parameters of the distribution (e.g., the range for a uniform distribution, μ and σ for a normal distribution, and the mean θ for an exponential distribution).

  4. Step 4 Specify the value x of interest in the “Input constant” box.

  5. Step 5 Click “OK”. The probability for the value of x will appear on the MINITAB session window.

    Figure 4.M.3

    MINITAB menu options for continuous probabilities

    [Note: Figure 4.M.4 gives the specifications for finding P(x23) in a normal distribution with μ=24.5 and σ=1

    Figure 4.M.4

    MINITAB normal distribution dialog box

Normal Probability Plot

  1. Step 1 Select “Graph” on the MINITAB menu bar, and then click on “Probability Plot,” as shown in Figure 4.M.5.

    Figure 4.M.5

    MINITAB options for a normal probability plot

  2. Step 2 Select “Single” (for one variable) on the next box, and the dialog box will appear as shown in Figure 4.M.6.

    Figure 4.M.6

    MINITAB normal probability plot dialog box

  3. Step 3 Specify the variable of interest in the “Graph variables” box, and then click the “Distribution” button and select the “Normal” option. Click “OK” to return to the Probability Plot dialog box.

  4. Step 4 Click “OK” to generate the normal probability plot.

TI-83/TI-84 Plus Graphing Calculator: Discrete Random Variables, Binomial, and Normal Probabilities

Calculating the Mean and Standard Deviation of a Discrete Random Variable

  1. Step 1 Enter the data

    • Press STAT and select 1:Edit

      (Note: If the lists already contain data, clear the old data. Use the up ARROW to highlight ‘L1.’)

    • Press CLEAR ENTER.

    • Use the up ARROW to highlight ‘L2

    • Press CLEAR ENTER

    • Use the ARROW and ENTER keys to enter the x-values of the variable into L1

    • Use the ARROW and ENTER keys to enter the probabilities, P(x), into L2

  2. Step 2 Access the Calc Menu

    • Press STAT

    • Arrow right to CALC

    • Select 1-Var Stats

    • Press ENTER

    • Press 2nd 1 for L1

    • Press COMMA

    • Select 2nd 2 for L2

    • Press ENTER

    The mean and standard deviation will be displayed on the screen, as will the quartiles, min, and max.

Calculating Binomial Probabilities

I. P(x=k)

To compute the probability of k successes in n trials where p is the probability of success for each trial, use the binompdf( command. Binompdf stands for “binomial probability density function.” This command is under the DISTRibution menu and has the format binompdf(n, p, k).

Example Compute the probability of 5 successes in 8 trials where the probability of success for a single trial is 40%. In this example, n=8,p=.4, and k=5.

  1. Step 1 Enter the binomial parameter

    • Press 2nd VARS for DISTR

    • Press the down ARROW key until binompdf is highlighted

    • Press ENTER

    • After binompdf(, type 8, .4, 5) (Note: Be sure to use the COMMA key between each parameter)

    • Press ENTER

    • You should see

    Thus, P(x=k) is about 12.4%.

II. P(xk)

To compute the probability of k or fewer successes in n trials where the p is probability of success for each trial, use the binomcdf( command. Binomcdf stands for “binomial cumulative probability density function.” This command is under the DISTRibution menu and has the format binomcdf(n, p, k).

Example Compute the probability of 5 or fewer successes in 8 trials where the probability of success for a single trial is 40%. In this example, n=8,p=.4, and k=5.

  1. Step 2 Enter the binomial parameters

    • Press 2nd VARS for DISTR

    • Press down the ARROW key until binomcdf is highlighted

    • Press ENTER

    • After binomcdf(, type 8, .4, 5)

    • Press ENTER

    • You should see

    Thus, P(x5) is about 95%.

III. P(x<k),P(x>k),P(xk)

To find the probability of less than k successes P(x<k), more than k successes P(x>k), or at least k successes P(xk), variations of the binomcdf( command must be used as shown below.

  1. P(x<k) use binomcdf(n, p, k1)

  2. P(x>k) use 1binomcdf(n, p, k)

  3. P(xk) use 1binomcdf(n, p, k1)

Graphing the Area under the Standard Normal Curve

  1. Step 1 Turn off all plots

    • Press 2nd PRGM and select 1:ClrDraw

    • Press ENTER, and ‘Done’ will appear on the screen

    • Press 2nd Y= and select 4:PlotsOff

    • Press ENTER, and ‘Done’ will appear on the screen

  2. Step 2 Set the viewing window (Recall that almost all of the area under the standard normal curve falls between 5 and 5. A height of 0.5 is a good choice for Ymax.)

    Note: When entering a negative number, be sure to use the negative sign ), and not the minus sign.

    • SeXmin=5

    • Xmax=5

    • Xscl=1

    • Ymin=0

    • Ymax=.5

    • Yscl=0

    • Xres=1

  3. Step 3 View graph

    • Press 2nd VARS

    • Arrow right to DRAW

    • Press ENTER to select 1:ShadeNorm(

    • Enter your lower limit (e.g., 5)

    • Press COMMA

    • Enter your upper limit (e.g., 1.5)

    • Press)

    • Press ENTER

    • The graph will be displayed along with the area, lower limit, and upper limit.

    • Thus, P(Z<1.5)=.9332

Finding Normal Probabilities without a Graph

To compute probabilities for a normal distribution, use the normalcdf( command. Normalcdf stands for “normal cumulative density function.” This command is under the DISTRibution menu and has the format normalcdf(lower limit, upper limit, mean, standard deviation).

  1. Step 1 Find the probability

    • Press 2nd VARS for DISTR and select Normalcdf(

    • After Normalcdf(, type in the lower limit

    • Press COMMA

    • Enter the upper limit

    • Press COMMA

    • Enter the mean

    • Press COMMA

    • Enter the standard deviation

    • Press)

    • Press ENTER

    The probability will be displayed on the screen.

    Example

    What is P(x<115) for a normal distribution with μ=100 and σ=10? In this example, the lower limit is , the upper limit is 115, the mean is 100, and the standard deviation is 10. To represent on the calculator, enter ) 1, press 2nd and press the COMMA key for EE, and then press 99. The screen appears as follows:

    • Thus, P(x<115)=.9332

Finding Normal Probabilities with a Graph

  1. Step 1 Turn off all plots

    • Press Y= and CLEAR all functions from the Y registers

    • Press 2nd Y= and select 4:PlotsOff

    • Press ENTER ENTER, and ‘Done’ will appear on the screen

  2. Step 2 Set the viewing window (These values depend on the mean and standard deviation of the data.) Note: When entering a negative number, be sure to use the negative sign ), not the minus sign.

    • Press WINDOW

    • SetXmin=μ5σ

    • Xmax=μ+5σ

    • Xscl=σ

    • Ymin=.125/σ

    • Ymax=.5/σ

    • Yscl=1

    • Xres=1

  3. Step 3 View graph

    • Press 2nd VARS

    • ARROW right to DRAW

    • Press ENTER to select 1:ShadeNorm(

    • Enter the lower limit

    • Press COMMA

    • Enter the upper limit

    • Press COMMA

    • Enter the mean

    • Press COMMA

    • Enter the standard deviation

    • Press)

    • Press ENTER

      The graph will be displayed along with the area, lower limit, and upper limit.

Example

What is P(x<115) for a normal distribution with μ=100 and σ=10? In this example, the lower limit is , the upper limit is 115, the mean is 100, and the standard deviation is 10. To represent on the calculator, enter ) 1, press 2nd and press the comma key for EE, and then press 99. The screens appear as follows:

Graphing a Normal Probability Plot

  1. Step 1 Enter the data

    • Press STAT and select 1:Edit

    • Note: If the list already contains data, clear the old data. Use the up arrow to highlight L1

    • Press CLEAR ENTER

    • Use the ARROW and ENTER keys to enter the data set into L1

  2. Step 2 Set up the normal probability plot

    • Press Y= and CLEAR all functions from the Y registers

    • Press 2nd and press Y= for STAT PLOT

    • Press 1 for Plot 1

    • Set the cursor so that ON is flashing

    • For Type, use the ARROW and ENTER keys to highlight and select the last graph in the bottom row

    • For Data List, choose the column containing the data (in most cases, L1)

      (Note: Press 2nd 1 for L1)

    • For Data Axis, choose X and press ENTER

  3. Step 3 View plot

    • Press ZOOM 9

Your data will be displayed against the expected z-scores from a normal distribution. If you see a “generally” linear relationship, your data set is approximately normal.

Simulating a Sampling Distribution

  1. Step 1 Select “Calc” on the MINITAB menu bar, and then click on “Random Data” (see Figure 4.M.7).

    Figure 4.M.7

    MINITAB options for generating random data

  2. Step 2 On the resulting menu list, click on the distribution of your choice (e.g., “Uniform”). A dialog box similar to the one (the Uniform Distribution) shown in Figure 4.M.8 will appear.

    Figure 4.M.8

    MINITAB dialog box for simulating the uniform distribution

  3. Step 3 Specify the number of samples (e.g., 1,000) to generate in the “Number of rows to generate” box and the columns where the data will be stored in the “Store in columns” box. (The number of columns will be equal to the sample size, e.g., 40.) Finally, specify the parameters of the distribution (e.g., the lower and upper range of the uniform distribution). When you click “OK,” the simulated data will appear on the MINITAB worksheet.

  4. Step 4 Calculate the value of the sample statistic of interest for each sample. To do this, click on the “Calc” button on the MINITAB menu bar, and then click on “Row Statistics,” as shown in Figure 4.M.9. The resulting dialog box appears in Figure 4.M.10.

    Figure 4.M.9

    MINITAB selections for generating sample statistics for the simulated data

    Figure 4.M.10

    MINITAB row statistics dialog box

  5. Step 5 Check the sample statistic (e.g., the mean) you want to calculate, specify the “Input variables” (or columns), and specify the column where you want the value of the sample statistic to be saved. Click “OK” and the value of the statistic for each sample will appear on the MINITAB worksheet.

[Note: Use the MINITAB menu choices provided in the Using Technology section in Chapter 2, pp. 112114, to generate a histogram of the sampling distribution of the statistic or to find the mean and variance of the sampling distribution.]

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