Textbook Question
Repeat Exercise 26 for samples of size 72 and 108. What happens to the mean and the standard deviation of the distribution of sample means as the sample size increases?
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Ch. 5 - Normal Probability Distributions
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470
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Table of contents
- Ch. 1 - Introduction to Statistics87
- Ch. 2 - Descriptive Statistics176
- Ch. 3 - Probability184
- Ch. 4 - Discrete Probability Distributions102
- Ch. 5 - Normal Probability Distributions183
- Ch. 6 - Confidence Intervals154
- Ch. 7 - Hypothesis Testing with One Sample165
- Ch. 8 - Hypothesis Testing with Two Samples134
- Ch. 9 - Correlation and Regression87
- Ch. 10 - Chi-Square Tests and the F-Distribution87
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 5, Problem 5.4.4
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 = 1275, sigma =6, n = 1000
Verified step by step guidance1
Step 1: Recall the formula for the mean of the sampling distribution of sample means. The mean of the sampling distribution (denoted as μₓ̄) is equal to the population mean (μ). Therefore, μₓ̄ = μ.
Step 2: Substitute the given value of the population mean (μ = 1275) into the formula. This means the mean of the sampling distribution is also 1275.
Step 3: Recall the formula for the standard deviation of the sampling distribution of sample means. The standard deviation of the sampling distribution (denoted as σₓ̄) is equal to the population standard deviation (σ) divided by the square root of the sample size (n). The formula is: σₓ̄ = σ / √n.
Step 4: Substitute the given values into the formula for σₓ̄. Use σ = 6 and n = 1000. This gives: σₓ̄ = 6 / √1000.
Step 5: Simplify the expression for σₓ̄ by calculating the square root of 1000 and dividing 6 by that value. This will give the standard deviation of 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
The sampling distribution of the sample mean is the probability distribution of all possible sample means from a population. It describes how the means of different samples will vary and is crucial for understanding the behavior of sample statistics. The Central Limit Theorem states that, regardless of the population's distribution, the sampling distribution will approach a normal distribution as the sample size increases.
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Sampling Distribution of Sample Proportion
Mean of the Sampling Distribution
The mean of the sampling distribution of the sample means, also known as the expected value, is equal to the population mean (mu). This means that if you take many samples and calculate their means, the average of those means will converge to the population mean. In this case, with mu = 1275, the mean of the sampling distribution will also be 1275.
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05:11Sampling Distribution of Sample Proportion
Standard Deviation of the Sampling Distribution
The standard deviation of the sampling distribution, known as the standard error, measures the dispersion of sample means around the population mean. It is calculated by dividing the population standard deviation (sigma) by the square root of the sample size (n). For this problem, with sigma = 6 and n = 1000, the standard error can be computed to understand how much variability to expect in the sample means.
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Related Practice
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 right of z= -0.355
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Textbook Question
Milk Containers A machine is set to fill milk containers with a mean of 64 ounces and a standard deviation of 0.11 ounce. A random sample of 40 containers has a mean of 64.05 ounces. The machine needs to be reset when the mean of a random sample is unusual. Does the machine need to be reset? Explain.
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Textbook Question
In Exercises 9–14, write the binomial probability in words. Then, use a continuity correction to convert the binomial probability to a normal distribution probability.
P(x ≥ 110)
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