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Summary

A normal distribution can be used to describe a variety of variables, such as heights, weights, and temperatures. A normal distribution is bell-shaped, unimodal, symmetric, and continuous; its mean, median, and mode are equal. Since each variable has its own distribution with mean μ and standard deviation σ, mathematicians use the standard normal distribution, which has a mean of 0 and a standard deviation of 1. Other approximately normally distributed variables can be transformed to the standard normal distribution with the formula z = (Xμ)/σ.

A normal distribution can also be used to describe a sampling distribution of sample means. These samples must be of the same size and randomly selected with replacement from the population. The means of the samples will differ somewhat from the population mean, since samples are generally not perfect representations of the population from which they came. The mean of the sample means will be equal to the population mean; and the standard deviation of the sample means will be equal to the population standard deviation, divided by the square root of the sample size. The central limit theorem states that as the size of the samples increases, the distribution of sample means will be approximately normal.

A normal distribution can be used to approximate other distributions, such as a binomial distribution. For a normal distribution to be used as an approximation, the conditions np ≥ 5 and nq ≥ 5 must be met. Also, a correction for continuity may be used for more accurate results.

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Important Formulas

Formula for the z value (or standard score):

Formula for finding a specific data value:

Formula for the mean of the sample means:

Formula for the standard error of the mean:

Formula for the z value for the central limit theorem:

Formulas for the mean and standard deviation for the binomial distribution:

problem-questionReview Exercises
  1. Find the area under the standard normal distribution curve for each.

    1. Between z = 0 and z = 1.95

    2. Between z = 0 and z = 0.37

    3. Between z = 1.32 and z = 1.82

    4. Between z = −1.05 and z = 2.05

    5. Between z = −0.03 and z = 0.53

    6. Between z = +1.10 and z = −1.80

    7. To the right of z = 1.99

    8. To the right of z = −1.36

    9. To the left of z = −2.09

    10. To the left of z = 1.68

  2. Using the standard normal distribution, find each probability.

    1. P(0 < z < 2.07)

    2. P(−1.83 < z < 0)

    3. P(−1.59 < z < +2.01)

    4. P(1.33 < z < 1.88)

    5. P(−2.56 < z < 0.37)

    6. P(z > 1.66)

    7. P(z < −2.03)

    8. P(z > −1.19)

    9. P(z < 1.93)

    10. P(z > −1.77)

  3. Per Capita Spending on Health Care The average per capita spending on health care in the United States is $5274. If the standard deviation is $600 and the distribution of health care spending is approximately normal, what is the probability that a randomly selected person spends more than $6000? Find the limits of the middle 50% of individual health care expenditures.

    Source: World Almanac.

  4. Salaries for Actuaries The average salary for graduates entering the actuarial field is $40,000. If the salaries are normally distributed with a standard deviation of $5000, find the probability that

    1. An individual graduate will have a salary over $45,000.

    2. A group of nine graduates will have a group average over $45,000.

    Source: www.BeAnActuary.org

  5. Speed Limits The speed limit on Interstate 75 around Findlay, Ohio, is 65 mph. On a clear day with no construction, the mean speed of automobiles was measured at 63 mph with a standard deviation of 8 mph. If the speeds are normally distributed, what percentage of the automobiles are exceeding the speed limit? If the Highway Patrol decides to ticket only motorists exceeding 72 mph, what percentage of the motorists might they arrest?

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  7. Monthly Spending for Paging and Messaging Services The average individual monthly spending in the United States for paging and messaging services is $10.15. If the standard deviation is $2.45 and the amounts are normally distributed, what is the probability that a randomly selected user of these services pays more than $15.00 per month? Between $12.00 and $14.00 per month?

    Source: New York Times Almanac.

  8. Average Precipitation For the first 7 months of the year, the average precipitation in Toledo, Ohio, is 19.32 inches. If the average precipitation is normally distributed with a standard deviation of 2.44 inches, find these probabilities.

    1. A randomly selected year will have precipitation greater than 18 inches for the first 7 months.

    2. Five randomly selected years will have an average precipitation greater than 18 inches for the first 7 months.

    Source: Toledo Blade.

  9. Suitcase Weights The average weight of an airline passenger’s suitcase is 45 pounds. The standard deviation is 2 pounds. If 15% of the suitcases are overweight, find the maximum weight allowed by the airline. Assume the variable is normally distributed.

  10. Confectionary Products Americans ate an average of 25.7 pounds of confectionary products each last year and spent an average of $61.50 per person doing so. If the standard deviation for consumption is 3.75 pounds and the standard deviation for the amount spent is $5.89, find the following:

    1. The probability that the sample mean confectionary consumption for a random sample of 40 American consumers was greater than 27 pounds.

    2. The probability that for a random sample of 50, the sample mean for confectionary spending exceeded $60.00.

    Source: www.census.gov

  11. Retirement Income Of the total population of American households, including older Americans and perhaps some not so old, 17.3% receive retirement income. In a random sample of 120 households, what is the probability that greater than 20 households but less than 35 households receive a retirement income?

    Source: www.bls.gov

  12. Portable CD Player Lifetimes A recent study of the life span of portable compact disc players found the average to be 3.7 years with a standard deviation of 0.6 year. If a random sample of 32 people who own CD players is selected, find the probability that the mean lifetime of the sample will be less than 3.4 years. If the mean is less than 3.4 years, would you consider that 3.7 years might be incorrect?

  13. Slot Machines The probability of winning on a slot machine is 5%. If a person plays the machine 500 times, find the probability of winning 30 times. Use the normal approximation to the binomial distribution.

  14. Multiple-Job Holders According to the government 5.3% of those employed are multiple-job holders. In a random sample of 150 people who are employed, what is the probability that fewer than 10 hold multiple jobs? What is the probability that more than 50 are not multiple-job holders?

    Source: www.bls.gov

  15. Enrollment in Personal Finance Course In a large university, 30% of the incoming first-year students elect to enroll in a personal finance course offered by the university. Find the probability that of 800 randomly selected incoming first-year students, at least 260 have elected to enroll in the course.

  16. U.S. Population Of the total population of the United States, 20% live in the northeast. If 200 residents of the United States are selected at random, find the probability that at least 50 live in the northeast.

    Source: Statistical Abstract of the United States.

  17. Heights of Active Volcanoes The heights (in feet above sea level) of a random sample of the world’s active volcanoes are shown here. Check for normality.

    13,435

    5,135

    11,339

    12,224

    7,470

    9,482

    12,381

    7,674

    5,223

    5,631

    3,566

    7,113

    5,850

    5,679

    15,584

    5,587

    8,077

    9,550

    8,064

    2,686

    5,250

    6,351

    4,594

    2,621

    9,348

    6,013

    2,398

    5,658

    2,145

    3,038

    Source: New York Times Almanac.
  18. Private Four-Year College Enrollment A random sample of enrollments in Pennsylvania’s private four-year colleges is listed here. Check for normality.

    1350

    1886

    1743

    1290

    1767

    2067

    1118

    3980

    1773

    4605

    1445

    3883

    1486

    980

    1217

    3587

     

     

     

     

    Source: New York Times Almanac.
  19. Construct a set of at least 15 data values which appear to be normally distributed. Verify the normality by using one of the methods introduced in this text.

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Statistics TodayWhat Is Normal?–Revisited

Many of the variables measured in medical tests—blood pressure, triglyceride level, etc.—are approximately normally distributed for the majority of the population in the United States. Thus, researchers can find the mean and standard deviation of these variables. Then, using these two measures along with the z values, they can find normal intervals for healthy individuals. For example, 95% of the systolic blood pressures of healthy individuals fall within 2 standard deviations of the mean. If an individual’s pressure is outside the determined normal range (either above or below), the physician will look for a possible cause and prescribe treatment if necessary.

Chapter Quiz

Determine whether each statement is true or false. If the statement is false, explain why.

  1. The total area under a normal distribution is infinite.

  2. The standard normal distribution is a continuous distribution.

  3. All variables that are approximately normally distributed can be transformed to standard normal variables.

  4. The z value corresponding to a number below the mean is always negative.

  5. The area under the standard normal distribution to the left of z = 0 is negative.

  6. The central limit theorem applies to means of samples selected from different populations.

Select the best answer.

  1. The mean of the standard normal distribution is

    • 0

    • b.

      1

    • c.

      100

    • d.

      Variable

  2. Approximately what percentage of normally distributed data values will fall within 1 standard deviation above or below the mean?

    • 68%

    • b.

      95%

    • c.

      99.7%

    • d.

      Variable

  3. Which is not a property of the standard normal distribution?

    • a.

      It’s symmetric about the mean.

    • It’s uniform.

    • c.

      It’s bell-shaped.

    • d.

      It’s unimodal.

  4. When a distribution is positively skewed, the relationship of the mean, median, and mode from left to right will be

    • a.

      Mean, median, mode

    • Mode, median, mean

    • c.

      Median, mode, mean

    • d.

      Mean, mode, median

  5. The standard deviation of all possible sample means equals

    • a.

      The population standard deviation

    • b.

      The population standard deviation divided by the population mean

    • The population standard deviation divided by the square root of the sample size

    • d.

      The square root of the population standard deviation

Complete the following statements with the best answer.

  1. When one is using the standard normal distribution, P(z < 0) =________.

  2. The difference between a sample mean and a population mean is due to ________.

  3. The mean of the sample means equals ________.

  4. The standard deviation of all possible sample means is called ________.

  5. The normal distribution can be used to approximate the binomial distribution when n·p and n·q are both greater than or equal to ________.

  6. The correction factor for the central limit theorem should be used when the sample size is greater than ________ the size of the population.

  7. Find the area under the standard normal distribution for each.

    1. Between 0 and 1.50

    2. Between 0 and −1.25

    3. Between 1.56 and 1.96

    4. Between −1.20 and −2.25

    5. Between −0.06 and 0.73

    6. Between 1.10 and −1.80

    7. To the right of z = 1.75

    8. To the right of z = −1.28

    9. To the left of z = −2.12

    10. To the left of z = 1.36

  8. Using the standard normal distribution, find each probability.

    1. P(0 < z < 2.16)

    2. P(−1.87 < z < 0)

    3. P(−1.63 < z < 2.17)

    4. P(1.72 < z < 1.98)

    5. P(−2.17 < z < 0.71)

    6. P(z > 1.77)

    7. P(z < −2.37)

    8. P(z > −1.73)

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    10. P(z < 2.03)

    11. P(z > −1.02)

  9. Amount of Rain in a City The average amount of rain per year in Greenville is 49 inches. The standard deviation is 8 inches. Find the probability that next year Greenville will receive the following amount of rainfall. Assume the variable is normally distributed.

    1. At most 55 inches of rain

    2. At least 62 inches of rain

    3. Between 46 and 54 inches of rain

    4. How many inches of rain would you consider to be an extremely wet year?

  10. Heights of People The average height of a certain age group of people is 53 inches. The standard deviation is 4 inches. If the variable is normally distributed, find the probability that a selected individual’s height will be

    1. Greater than 59 inches

    2. Less than 45 inches

    3. Between 50 and 55 inches

    4. Between 58 and 62 inches

  11. Lemonade Consumption The average number of gallons of lemonade consumed by the football team during a game is 20, with a standard deviation of 3 gallons. Assume the variable is normally distributed. When a game is played, find the probability of using

    1. Between 20 and 25 gallons

    2. Less than 19 gallons

    3. More than 21 gallons

    4. Between 26 and 28 gallons

  12. Years to Complete a Graduate Program The average number of years a person takes to complete a graduate degree program is 3. The standard deviation is 4 months. Assume the variable is normally distributed. If an individual enrolls in the program, find the probability that it will take

    1. More than 4 years to complete the program

    2. Less than 3 years to complete the program

    3. Between 3.8 and 4.5 years to complete the program

    4. Between 2.5 and 3.1 years to complete the program

  13. Passengers on a Bus On the daily run of an express bus, the average number of passengers is 48. The standard deviation is 3. Assume the variable is normally distributed. Find the probability that the bus will have

    1. Between 36 and 40 passengers

    2. Fewer than 42 passengers

    3. More than 48 passengers

    4. Between 43 and 47 passengers

  14. Thickness of Library Books The average thickness of books on a library shelf is 8.3 centimeters. The standard deviation is 0.6 centimeter. If 20% of the books are oversized, find the minimum thickness of the oversized books on the library shelf. Assume the variable is normally distributed.

  15. Membership in an Organization Membership in an elite organization requires a test score in the upper 30% range. If μ = 115 and σ= 12, find the lowest acceptable score that would enable a candidate to apply for membership. Assume the variable is normally distributed.

  16. Repair Cost for Microwave Ovens The average repair cost of a microwave oven is $55, with a standard deviation of $8. The costs are normally distributed. If 12 ovens are repaired, find the probability that the mean of the repair bills will be greater than $60.

  17. Electric Bills The average electric bill in a residential area is $72 for the month of April. The standard deviation is $6. If the amounts of the electric bills are normally distributed, find the probability that the mean of the bill for 15 residents will be less than $75.

  18. Sleep Survey According to a recent survey, 38% of Americans get 6 hours or less of sleep each night. If 25 people are selected, find the probability that 14 or more people will get 6 hours or less of sleep each night. Does this number seem likely?

    Source: Amazing Almanac.

  19. Factory Union Membership If 10% of the people in a certain factory are members of a union, find the probability that, in a sample of 2000, fewer than 180 people are union members.

  20. Household Online Connection The percentage of U.S. households that have online connections is 44.9%. In a random sample of 420 households, what is the probability that fewer than 200 have online connections?

    Source: New York Times Almanac.

  21. Computer Ownership Fifty-three percent of U.S. households have a personal computer. In a random sample of 250 households, what is the probability that fewer than 120 have a PC?

    Source: New York Times Almanac.

  22. Calories in Fast-Food Sandwiches The number of calories contained in a selection of fast-food sandwiches is shown here. Check for normality.

    390

    405

    580

    300

    320

    540

    225

    720

    470

    560

    535

    660

    530

    290

    440

    390

    675

    530

    1010

    450

    320

    460

    290

    340

    610

    430

    530

     

     

     

    Source: The Doctor’s Pocket Calorie, Fat, and Carbohydrate Counter.
  23. GMAT Scores The average GMAT scores for the top-30 ranked graduate schools of business are listed here. Check for normality.

    718

    703

    703

    703

    700

    690

    695

    705

    690

    688

    676

    681

    689

    686

    691

    669

    674

    652

    680

    670

    651

    651

    637

    662

    641

    645

    645

    642

    660

    636

    Source: U.S. News & World Report Best Graduate Schools.
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Critical Thinking Challenges

Sometimes a researcher must decide whether a variable is normally distributed. There are several ways to do this. One simple but very subjective method uses special graph paper, which is called normal probability paper. For the distribution of systolic blood pressure readings given in Chapter 3 of the textbook, the following method can be used:

  1. Make a table, as shown.

    Boundaries

    Frequency

    Cumulative frequency

    Cumulative percent frequency

      89.5–104.5

    24

     

     

    104.5–119.5

    62

     

     

    119.5–134.5

    72

     

     

    134.5–149.5

    26

     

     

    149.5–164.5

    12

     

     

    164.5–179.5

      4

     

     

     

    200

     

     

  2. Find the cumulative frequencies for each class, and place the results in the third column.

  3. Find the cumulative percents for each class by dividing each cumulative frequency by 200 (the total frequencies) and multiplying by 100%. (For the first class, it would be 24/200 × 100% = 12%.) Place these values in the last column.

  4. Using the normal probability paper shown in Table 6–3, label the x axis with the class boundaries as shown and plot the percents.

    Table 6-3
    Normal Probability Paper
  5. If the points fall approximately in a straight line, it can be concluded that the distribution is normal. Do you feel that this distribution is approximately normal? Explain your answer.

  6. To find an approximation of the mean or median, draw a horizontal line from the 50% point on the y axis over to the curve and then a vertical line down to the x axis. Compare this approximation of the mean with the computed mean.

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  8. To find an approximation of the standard deviation, locate the values on the x axis that correspond to the 16 and 84% values on the y axis. Subtract these two values and divide the result by 2. Compare this approximate standard deviation to the computed standard deviation.

  9. Explain why the method used in step 7 works.

Data Projects
  1. Business and Finance Use the data collected in data project 1 of Chapter 2 regarding earnings per share to complete this problem. Use the mean and standard deviation computed in data project 1 of Chapter 3 as estimates for the population parameters. What value separates the top 5% of stocks from the others?

  2. Sports and Leisure Find the mean and standard deviation for the batting average for a player in the most recently completed MBL season. What batting average would separate the top 5% of all hitters from the rest? What is the probability that a randomly selected player bats over 0.300? What is the probability that a team of 25 players has a mean that is above 0.275?

  3. Technology Use the data collected in data project 3 of Chapter 2 regarding song lengths. If the sample estimates for mean and standard deviation are used as replacements for the population parameters for this data set, what song length separates the bottom 5% and top 5% from the other values?

  4. Health and Wellness Use the data regarding heart rates collected in data project 4 of Chapter 2 for this problem. Use the sample mean and standard deviation as estimates of the population parameters. For the before-exercise data, what heart rate separates the top 10% from the other values? For the after-exercise data, what heart rate separates the bottom 10% from the other values? If a student was selected at random, what is the probability that her or his mean heart rate before exercise was less than 72? If 25 students were selected at random, what is the probability that their mean heart rate before exercise was less than 72?

  5. Politics and Economics Use the data collected in data project 6 of Chapter 2 regarding Math SAT scores to complete this problem. What are the mean and standard deviation for statewide Math SAT scores? What SAT score separates the bottom 10% of states from the others? What is the probability that a randomly selected state has a statewide SAT score above 500?

  6. Your Class Confirm the two formulas hold true for the central limit theorem for the population containing the elements {1, 5, 10}. First, compute the population mean and standard deviation for the data set. Next, create a list of all 9 of the possible two-element samples that can be created with replacement: {1, 1}, {1, 5}, etc. For each of the 9 compute the sample mean. Now find the mean of the sample means. Does it equal the population mean? Compute the standard deviation of the sample means. Does it equal the population standard deviation, divided by the square root of n?

Answers to Applying the Concepts
Section 6-1
Assessing Normality
  1. Answers will vary. One possible frequency distribution is the following:

    Branches

    Frequency

    0–9

      1

    10–19

    14

    20–29

    17

    30–39

      7

    40–49

      3

    50–59

      2

    60–69

      2

    70–79

      1

    80–89

      2

    90–99

      1

  2. Answers will vary according to the frequency distribution in question 1. This histogram matches the frequency distribution in question 1.

  3. The histogram is unimodal and skewed to the right (positively skewed).

  4. The distribution does not appear to be normal.

  5. Page 354
  6. The mean number of branches is , and the standard deviation is s = 20.6.

  7. Of the data values, 80% fall within 1 standard deviation of the mean (between 10.8 and 52).

  8. Of the data values, 92% fall within 2 standard deviations of the mean (between 0 and 72.6).

  9. Of the data values, 98% fall within 3 standard deviations of the mean (between 0 and 93.2).

  10. My values in questions 6–8 differ from the 68, 95, and 100% that we would see in a normal distribution.

  11. These values support the conclusion that the distribution of the variable is not normal.

Section 6-2
Smart People
  1. . The area to the right of 2 in the standard normal table is about 0.0228, so I would expect about 10,000(0.0228) = 228 people in Visiala to qualify for Mensa.

  2. It does seem reasonable to continue my quest to start a Mensa chapter in Visiala.

  3. Answers will vary. One possible answer would be to randomly call telephone numbers (both home and cell phones) in Visiala, ask to speak to an adult, and ask whether the person would be interested in joining Mensa.

  4. To have an Ultra-Mensa club, I would need to find the people in Visiala who have IQs that are at least 2.326 standard deviations above average. This means that I would need to recruit those with IQs that are at least 135:

Section 6-3
Central Limit Theorem
  1. It is very unlikely that we would ever get the same results for any of our random samples. While it is a remote possibility, it is highly unlikely.

  2. A good estimate for the population mean would be to find the average of the students’ sample means. Similarly, a good estimate for the population standard deviation would be to find the average of the students’ sample standard deviations.

  3. The distribution appears to be somewhat left (negatively) skewed.

  4. The mean of the students’ means is 25.4, and the standard deviation is 5.8.

  5. The distribution of the means is not a sampling distribution, since it represents just 20 of all possible samples of size 30 from the population.

  6. The sampling error for student 3 is 18 − 25.4 = −7.4; the sampling error for student 7 is 26 − 25.4 = +0.6; the sampling error for student 14 is 29 − 25.4 = +3.6.

  7. The standard deviation for the sample of the 20 means is greater than the standard deviations for each of the individual students. So it is not equal to the standard deviation divided by the square root of the sample size.

Section 6-4
How Safe Are You?
  1. A reliability rating of 97% means that, on average, the device will not fail 97% of the time. We do not know how many times it will fail for any particular set of 100 climbs.

  2. The probability of at least 1 failure in 100 climbs is 1− (0.97)100 = 1 − 0.0476 = 0.9524 (about 95%).

  3. The complement of the event in question 2 is the event of “no failures in 100 climbs.”

  4. This can be considered a binomial experiment. We have two outcomes: success and failure. The probability of the equipment working (success) remains constant at 97%. We have 100 independent climbs. And we are counting the number of times the equipment works in these 100 climbs.

  5. We could use the binomial probability formula, but it would be very messy computationally.

  6. The probability of at least two failures cannot be estimated with the normal distribution (see below). So the probability is 1 − [(0.97)100 + 100(0.97)99 (0.03)] =1 − 0.1946 = 0.8054 (about 80.5%).

  7. We should not use the normal approximation to the binomial since nq < 10.

  8. If we had used the normal approximation, we would have needed a correction for continuity, since we would have been approximating a discrete distribution with a continuous distribution.

  9. Since a second safety hook will be successful or fail independently of the first safety hook, the probability of failure drops from 3% to (0.03)(0.03) = 0.0009, or 0.09%.