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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 5.4.9
Chapter 5, Problem 5.4.9

Graphical Analysis In Exercises 9 and 10, the graph of a population distribution is shown with its mean and standard deviation. Random samples of size 100 are drawn from the population. Determine which of the figures labeled (a)–(c) would most closely resemble the sampling distribution of sample means. Explain your reasoning.


The waiting time (in seconds) to turn left at an intersection

Graph showing a bell curve with mean at 16.5 seconds and standard deviation at 11.9 seconds, labeled on the x-axis.

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Step 1: Understand the problem. The question asks us to determine which figure (a, b, or c) most closely resembles the sampling distribution of sample means for random samples of size 100 drawn from the population. The population distribution is shown in the first graph, with a mean (μ) of 16.5 seconds and a standard deviation (σ) of 11.9 seconds.
Step 2: Recall the Central Limit Theorem. According to the theorem, the sampling distribution of the sample mean will be approximately normal if the sample size is sufficiently large (n ≥ 30). Since the sample size is 100, the sampling distribution of the sample mean will be normal regardless of the shape of the population distribution.
Step 3: Calculate the mean and standard deviation of the sampling distribution. The mean of the sampling distribution (μₓ̄) is equal to the population mean (μ), which is 16.5 seconds. The standard deviation of the sampling distribution (σₓ̄) is equal to the population standard deviation (σ) divided by the square root of the sample size (n). Use the formula: σx=σn.
Step 4: Compare the graphs. The first graph represents the population distribution, which is skewed to the right. The second graph represents a normal distribution with the same mean (16.5 seconds) and a smaller standard deviation, which matches the characteristics of the sampling distribution of sample means.
Step 5: Conclude that the second graph (b) most closely resembles the sampling distribution of sample means. This is because the sampling distribution is normal due to the Central Limit Theorem, and its mean and standard deviation align with the calculated values for the sampling distribution.

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Key Concepts

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

Sampling Distribution of Sample Means

The sampling distribution of sample means is the probability distribution of all possible sample means from a population. When random samples of size 100 are drawn, the Central Limit Theorem states that this distribution will tend to be normally distributed, regardless of the population's distribution, especially as the sample size increases. This concept is crucial for understanding how sample means behave and how they can be used to make inferences about the population mean.
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Central Limit Theorem (CLT)

The Central Limit Theorem is a fundamental statistical principle that states that the distribution of the sample means will approach a normal distribution as the sample size becomes large, typically n ≥ 30. This theorem allows statisticians to make inferences about population parameters using sample statistics, as it provides a basis for the normal approximation of the sampling distribution, even if the original population distribution is not normal.
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Bayes' Theorem

Mean and Standard Deviation

The mean (μ) is the average of a set of values, representing the central point of a distribution, while the standard deviation (σ) measures the dispersion or spread of the values around the mean. In the context of the question, the mean waiting time is 16.5 seconds, and the standard deviation is 11.9 seconds. These parameters are essential for understanding the characteristics of the population distribution and how they influence the shape of the sampling distribution of sample means.
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Related Practice
Textbook Question

Graphical Analysis In Exercises 17–22, find the indicated z-score(s) shown in the graph.


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Textbook Question

Finding Probabilities In Exercises 15–18, the population mean and standard deviation are given. Find the indicated probability and determine whether the given sample mean would be considered unusual.


For a random sample of n=36, find the probability of a sample mean being less than 12,750 or greater than 12,753 when mu=12750 and 1.7.

Textbook Question

Computing and Interpreting z-Scores In Exercises 39 and 40, (a) find the z-score that corresponds to each value and (b) determine whether any of the values are unusual.


Stanford-Binet IQ Scores The test scores for the Stanford-Binet Intelligence Scale are normally distributed with a mean score of 100 and a standard deviation of 16. The test scores of four students selected at random are 98, 65, 106, and 124.

Textbook Question

Finding Area

In Exercises 23–36, find the indicated area under the standard normal curve. If convenient, use technology to find the area.


To the left of z=0.33

Textbook Question

In Exercises 1–4, a population has a mean mu and a standard deviation sigma. Find the mean and standard deviation of the sampling distribution of sample means with sample size n.


Mu = 150, sigma =25, n = 49

Textbook Question

Construction About 63% of the residents in a town are in favor of building a new high school. One hundred five residents are randomly selected. What is the probability that the sample proportion in favor of building a new school is less than 55%? Interpret your result.