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Ch. 5 - Normal Probability Distributions
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
All textbooksLarson 8th EditionCh. 5 - Normal Probability DistributionsProblem 5.2.17b
Chapter 5, Problem 5.2.17b

SAT Total Scores Use the normal distribution in Exercise 13.
b. Out of 1000 randomly selected SAT total scores, about how many would you expect to be greater than 1100?

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Identify the key components of the problem: The SAT total scores are normally distributed, and we are tasked with finding the number of scores greater than 1100 out of 1000 randomly selected scores. This requires calculating the probability of a score being greater than 1100 and then scaling it to the sample size of 1000.
Standardize the score of 1100 using the z-score formula: z = X - μσ, where X is the score (1100), μ is the mean of the distribution, and σ is the standard deviation. These values should be provided in the problem or assumed based on typical SAT score distributions (e.g., μ = 1000 and σ = 200).
Use the z-score to find the cumulative probability from the standard normal distribution table or a statistical software. This gives the probability of a score being less than 1100. Subtract this value from 1 to find the probability of a score being greater than 1100: P(X > 1100) = 1 - P(Z ≤ z).
Multiply the probability of a score being greater than 1100 by the total number of scores (1000) to find the expected number of scores greater than 1100: E = P(X > 1100) × 1000.
Interpret the result: The final value represents the expected number of SAT scores greater than 1100 out of the 1000 randomly selected scores.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Normal Distribution

The normal distribution is a probability distribution that is symmetric about the mean, depicting that data near the mean are more frequent in occurrence than data far from the mean. It is characterized by its bell-shaped curve and is defined by two parameters: the mean (average) and the standard deviation (which measures the spread of the data). Understanding this distribution is crucial for making inferences about populations based on sample data.
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Z-scores

A Z-score indicates how many standard deviations an element is from the mean of a distribution. It is calculated by subtracting the mean from the value and then dividing by the standard deviation. Z-scores are essential for determining the probability of a score occurring within a normal distribution, allowing us to find the proportion of scores that fall above or below a certain value, such as 1100 in this case.
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Empirical Rule

The empirical rule, also known as the 68-95-99.7 rule, states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations. This rule helps in estimating the likelihood of scores exceeding a certain threshold, such as 1100, by providing a quick way to assess how many scores lie beyond a specified number of standard deviations from the mean.
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Related Practice
Textbook Question

Finding Specified Data Values In Exercises 31–38, answer the questions about the specified normal distribution.


Water Footprint A water footprint is a measure of the appropriation of fresh water. The water footprint (in cubic meters) for a kilogram of wheat can be approximated by a normal distribution, as shown in the figure. (Source: Ecological Indicators)



b. What water footprint represents the 29th percentile?

Textbook Question

Red Blood Cell Count Use the normal distribution in Exercise 16.


a. What percent of the adult males have a red blood cell count less than 6 million cells per microliter?

Textbook Question

Approximating Binomial Probabilities In Exercises 19–26, determine whether you can use a normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you cannot, explain why and use a binomial distribution to find the indicated probabilities. Identify any unusual events. Explain.


Social Media A survey of Americans found that 55% would be disappointed if Facebook disappeared. You randomly select 500 Americans and ask them whether they would be disappointed if Facebook disappeared. Find the probability that the number who say yes is (a) less than 250

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Textbook Question

Finding Specified Data Values In Exercises 31–38, answer the questions about the specified normal distribution.


Advanced Dental Admission Test The Advanced Dental Admission Test (ADAT) is designed so that the scores fit a normal distribution, as shown in the figure. (Source: American Dental Association)



b. Between what two values does the middle 50% of the ADAT scores lie?

Textbook Question

Approximating Binomial Probabilities In Exercises 19–26, determine whether you can use a normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you cannot, explain why and use a binomial distribution to find the indicated probabilities. Identify any unusual events. Explain.


Advancing Research In a survey of U.S. adults, 77% said are willing to share their personal health information to advance medical research. You randomly select 500 U.S. adults. Find the probability that the number who are willing to share their personal health information to advance medical research is (b) more than 360

Textbook Question

Finding Probabilities for Normal Distributions In Exercises 7–12, find the indicated probabilities. If convenient, use technology to find the probabilities.

         

Health Club Schedule The amounts of time per workout an athlete uses a stairclimber are normally distributed, with a mean of 20 minutes and a standard deviation of 5 minutes. Find the probability that a randomly selected athlete uses a stairclimber for (a) less than 17 minutes.