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Ch. 5 - Normal Probability Distributions
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470당신이 사용하는 게 아니라요?교과서 변경
5장, 문제 5.4.24

Interpreting the Central Limit Theorem In Exercises 19–26, find the mean and standard deviation of the indicated sampling distribution of sample means. Then sketch a graph of the sampling distribution.


Salaries The annual salaries for web software development managers are normally distributed, with a mean of about \$136,000 and a standard deviation of about \$11,500. Random samples of 40 are drawn from this population, and the mean of each sample is determined.

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Identify the population mean (\$\(\mu\)\$) and population standard deviation (\$\(\sigma\)\$). Here, \$\(\mu\) = 136,000\$ and \$\(\sigma\) = 11,500\$.
Note the sample size \$n = 40\$ and recognize that the sampling distribution of the sample mean will have its own mean and standard deviation.
Calculate the mean of the sampling distribution of the sample mean, which is the same as the population mean: \$\(\mu\)_{\(\bar{x}\)} = \(\mu\) = 136,000\$.
Calculate the standard deviation of the sampling distribution of the sample mean (also called the standard error) using the formula: \$\(\sigma\)_{\(\bar{x}\)} = \(\frac{\sigma}{\sqrt{n}\)} = \(\frac{11,500}{\sqrt{40}\)}\$.
Since the population distribution is normal, the sampling distribution of the sample mean will also be normal. Sketch a normal curve centered at \$136,000\$ with spread determined by \$\(\sigma\)_{\(\bar{x}\)}\$.

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주요 개념

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Central Limit Theorem (CLT)

The Central Limit Theorem states that the sampling distribution of the sample mean approaches a normal distribution as the sample size becomes large, regardless of the population's distribution. For sample sizes typically greater than 30, the sample means will be approximately normally distributed, enabling inference about the population mean.
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가이드 코스
05:14
Central Limit Theorem

Sampling Distribution of the Sample Mean

The sampling distribution of the sample mean is the probability distribution of all possible sample means from samples of a fixed size. It has a mean equal to the population mean and a standard deviation (standard error) equal to the population standard deviation divided by the square root of the sample size.
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가이드 코스
06:53
Sampling Distribution of Sample Mean

Standard Error of the Mean

The standard error measures the variability of the sample mean from the population mean. It is calculated as the population standard deviation divided by the square root of the sample size, reflecting how much sample means fluctuate around the true mean in repeated sampling.
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가이드 코스
04:52
Calculating the Mean
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In Exercises 23–36, find the indicated area under the standard normal curve. If convenient, use technology to find the area.


Between z=0 and z=2.86

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


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교과서 질문

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.

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교과서 질문

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.

155
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교과서 질문

Computing Probabilities for Normal Distributions In Exercises 1–6, the random variable x is normally distributed with mean mu=174 and standard deviation sigma=20. Find the indicated probability.


P(172 < x <192)

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교과서 질문

Finding Probability In Exercises 47–56, find the indicated probability using the standard normal distribution. If convenient, use technology to find the probability.


P(- 1.54 < z < 1.54)

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