Chapter 6: Problem 12
A population has mean 12 and standard deviation \(1.5 .\) a. Find the mean and standard deviation of \(x-\) for samples of size 90 . b. Find the probability that the mean of a sample of size 90 will differ from the population mean 12 by at least 0.3 unit, that is, is either less than 11.7 or more than 12.3 .
Short Answer
Step by step solution
Understanding the Central Limit Theorem
Identify the Mean of the Sample Mean (x̄)
Identify the Standard Deviation of the Sample Mean (x̄)
Calculate the Z-scores for the Sample Mean
Calculate the Probability
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sampling Distribution
- The sampling distribution of the mean will have the same mean as the population mean.
- The variability or spread is less than that of the population, determined by the standard error.
Standard Deviation
- The formula for the standard error of the mean is: \[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\] where \(\sigma_{\bar{x}}\) is the standard error, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
- In our exercise, it was calculated as \(0.15857\) for a sample of size 90.
Probability Calculation
- We are interested in the probability that the sample mean falls outside of the specified range (11.7 to 12.3 in our exercise).
- This involves finding the probability for the corresponding z-scores (probabilities are the areas under the z-score curve).
Z-scores
- The formula to calculate the z-score is: \[Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}\] where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, and \(\sigma_{\bar{x}}\) is the standard error.
- In our context, we calculated z-scores for two sample means (11.7 and 12.3) and obtained values of approximately -1.89 and 1.89.