Solved: Intelligence quotients (IQs) on the Stanford-Binet

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is 1 standard deviation above the mean.

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is 2 standard deviations above the mean.

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is 3 standard deviations above the mean.

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is \(1 \frac{1}{2}\) standard deviations above the mean. Text Transcription: 1 frac 1 2

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is \(2 \frac{1}{2}\) standard deviations above the mean. Text Transcription: 2 frac 1 2

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is 1 standard deviation below the mean.

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is 2 standard deviations below the mean.

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is 3 standard deviations below the mean.

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is one-half a standard deviation below the mean.

The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. In Exercises 1–10, find the score that is \(2 \frac{1}{2}\) standard deviations below the mean. Text Transcription: 2 frac 1 2

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid between $16,500 and $17,500.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid between $16,000 and $18,000.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid between $17,000 and $17,500.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid between $17,000 and $18,000.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid between $16,000 and $17,000.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid between $16,500 and $17,000.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid between $15,500 and $17,000.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid between $17,000 and $18,500.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid more than $17,500.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid more than $18,000.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid less than $16,000.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. In Exercises 11–22, use the 68–95–99.7 Rule, illustrated in the figure, to find the percentage of buyers who paid less than $16,500.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs between 68 and 132.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs between 84 and 116.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs between 68 and 100.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs between 84 and 100.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs above 116.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs above 132.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs below 68.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs below 84.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs above 148.

Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $17,000 and the standard deviation is $500. Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. In Exercises 23–32, use the 68–95–99.7 Rule to find the percentage of people with IQs below 52.

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 68

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 76

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 84

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 92

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 64

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 72

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 74

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 78

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 60

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 100

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 52

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 44

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 48

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 40

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 34

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. In Exercises 33–48, convert each data item to a z-score. 30

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. In Exercises 49–56, find the z-score for the given score on this dental anxiety scale. 17

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. In Exercises 49–56, find the z-score for the given score on this dental anxiety scale. 18

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. In Exercises 49–56, find the z-score for the given score on this dental anxiety scale. 20

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. In Exercises 49–56, find the z-score for the given score on this dental anxiety scale. 12

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. In Exercises 49–56, find the z-score for the given score on this dental anxiety scale. 6

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. Find the z-score for the given score on this dental anxiety scale. 8

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. In Exercises 49–56, find the z-score for the given score on this dental anxiety scale. 5

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. In Exercises 49–56, find the z-score for the given score on this dental anxiety scale. 1

Intelligence quotients on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. Intelligence quotients on the Wechsler intelligence test are normally distributed with a mean of 100 and a standard deviation of 15. Use this information to solve Exercises 57–58. Use z-scores to determine which person has the higher IQ: an individual who scores 128 on the Stanford-Binet or an individual who scores 127 on the Wechsler.

Intelligence quotients on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. Intelligence quotients on the Wechsler intelligence test are normally distributed with a mean of 100 and a standard deviation of 15. Use this information to solve Exercises 57–58. Use z-scores to determine which person has the higher IQ: an individual who scores 150 on the Stanford-Binet or an individual who scores 148 on the Wechsler.

A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises 59–66, find the data item in this distribution that corresponds to the given z-score. z = 2

A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises 59–66, find the data item in this distribution that corresponds to the given z-score. z = 3

A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises 59–66, find the data item in this distribution that corresponds to the given z-score. z = 1.5

A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises 59–66, find the data item in this distribution that corresponds to the given z-score. z = 2.5

A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises 59–66, find the data item in this distribution that corresponds to the given z-score. z = -3

A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises 59–66, find the data item in this distribution that corresponds to the given z-score. z = -2

A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises 59–66, find the data item in this distribution that corresponds to the given z-score. z = -2.5

A set of data items is normally distributed with a mean of 400 and a standard deviation of 50. In Exercises 59–66, find the data item in this distribution that corresponds to the given z-score. z = -1.5

Reducing Gun Violence The data in the bar graph are from a random sample of 814 American adults. The graph shows four proposals to reduce gun violence in the United States and the percentage of surveyed adults who favored each of these proposals. a. Find the margin of error, to the nearest tenth of a percent, for this survey. b. Write a statement about the percentage of adults in the U.S population who favor required gun registration to reduce gun violence.

How to Blow Your Job Interview The data in the bar graph at the top of the next column are from a random sample of 1910 job interviewers. The graph shows the top interviewer turnoffs and the percentage of surveyed interviewers who were offended by each of these behaviors. a. Find the margin of error, to the nearest tenth of a percent, for this survey. b. Write a statement about the percentage of interviewers in the population who are turned off by a job applicant being arrogant.

Using a random sample of 4000 TV households, Nielsen Media Research found that 60.2% watched the final episode of M*A*S*H. a. Find the margin of error in this percent. b. Write a statement about the percentage of TV households in the population that tuned into the final episode of M*A*S*H.

Using a random sample of 4000 TV households, Nielsen Media Research found that 51.1% watched Roots, Part 8. a. Find the margin of error in this percent. b. Write a statement about the percentage of TV households in the population that tuned into Roots, Part 8.

In 1997, Nielsen Media Research increased its random sample to 5000 TV households. By how much, to the nearest tenth of a percent, did this improve the margin of error over that in Exercises 69 and 70?

If Nielsen Media Research were to increase its random sample from 5000 to 10,000 TV households, by how much, to the nearest tenth of a percent, would this improve the margin of error?

The histogram shows murder rates per 100,000 residents and the number of U.S. states that had these rates for a recent year. Use this histogram to solve Exercises 73–74. a. Is the shape of this distribution best classified as normal, skewed to the right, or skewed to the left? b. Calculate the mean murder rate per 100,000 residents for the 50 states and Washington, D.C. c. Find the median murder rate per 100,000 residents for the 50 states and Washington, D.C. d. Are the mean and median murder rates consistent with the shape of the distribution that you described in part (a)? Explain your answer. e. The standard deviation for the data is approximately 4.2. If the distribution were roughly normal, what would be the z-score, rounded to one decimal place, for Washington, D.C.? Does this seem unusually high? Explain your answer.

The histogram shows murder rates per 100,000 residents and the number of U.S. states that had these rates for a recent year. Use this histogram to solve Exercises 73–74. a. Find the median murder rate per 100,000 residents for the 50 states and Washington, D.C. b. Find the first quartile by determining the median of the lower half of the data. (This is the median of the items that lie below the median that you found in part (a).) c. Find the third quartile by determining the median of the upper half of the data. (This is the median of the items that lie above the median that you found in part (a).) d. Use the following numerical scale: Above this scale, show five points, each at the same height. (The height is arbitrary.) Each point should represent one of the following numbers: lowest data value, first quartile, median, third quartile, highest data value. e. A box-and-whisker plot consists of a rectangular box extending from the first quartile to the third quartile, with a dashed line representing the median, and line segments (or whiskers) extending outward from the box to the lowest and highest data values: Use your graph from part (d) to create a box-and-whisker plot for U.S. murder rates per 100,000 residents. f. If one of the whiskers in a box-and-whisker plot is clearly longer, the distribution is usually skewed in the direction of the longer whisker. Based on this observation, does your box-and-whisker plot in part (e) indicate that the distribution is skewed to the right or skewed to the left? g. Is the shape of the distribution of scores shown by the given histogram consistent with your observation in part (f)?

What is a symmetric histogram?

Describe the normal distribution and discuss some of its properties.

Describe the 689599.7 Rule

Describe how to determine the z-score for a data item in a normal distribution.

What does a z-score measure?

Give an example of both a commonly occurring and an infrequently occurring z-score. Explain how you arrived at these examples.

Describe when a z-score is negative

If you score in the 83rd percentile, what does this mean?

If your weight is in the third quartile, what does this mean?

Two students have scores with the same percentile, but for different administrations of the SAT. Does this mean that the students have the same score on the SAT? Explain your answer.

Give an example of a phenomenon that is normally distributed. Explain why. (Try to be creative and not use one of the distributions discussed in this section.) Estimate what the mean and the standard deviation might be and describe how you determined these estimates.

Give an example of a phenomenon that is not normally distributed and explain why.

In Exercises 87–90, determine whether each statement makes sense or does not make sense, and explain your reasoning. The heights of the men on our college basketball team are normally distributed with a mean of 6 feet 3 inches and a standard deviation of 1 foot 2 inches.

In Exercises 87–90, determine whether each statement makes sense or does not make sense, and explain your reasoning. I scored in the 50th percentile on a standardized test, so my score is the median.

In Exercises 87–90, determine whether each statement makes sense or does not make sense, and explain your reasoning. A poll administered to a random sample of 1150 voters shows 51% in favor of candidate A, so I’m 95% confident that candidate A will win the election.

In Exercises 87–90, determine whether each statement makes sense or does not make sense, and explain your reasoning. My math teacher gave a very difficult exam for which the distribution of scores was skewed to the right.

For this activity, group members will conduct interviews with a random sample of students on campus. Each student is to be asked, What is the worst thing about being a student? One response should be recorded for each student. a. Each member should interview enough students so that there are at least 50 randomly selected students in the sample. b. After all responses have been recorded, the group should organize the four most common answers. For each answer, compute the percentage of students in the sample who felt that this is the worst thing about being a student. c. Find the margin of error for your survey. d. For each of the four most common answers, write a statement about the percentage of all students on your campus who feel that this is the worst thing about being a student