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Chapter 8: Problem 12

Determine \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) from the given parameters of the population and the sample size. \(\mu=27, \sigma=6, n=15\)

Short Answer

Expert verified
\[ \mu_{\bar{x}} = 27 \] \[ \sigma_{\bar{x}} \approx 1.55 \]

Step by step solution

01

- Understand the Central Tendency and Spread

Given the population mean \( \mu = 27 \), standard deviation \( \sigma = 6 \), and sample size \( \ n = 15 \), determine the mean and the standard deviation of the sample mean.
02

- Calculate the Mean of the Sample Mean

The mean of the sample mean, \( \mu_{\bar{x}} \), is equal to the population mean. So, \[ \mu_{\bar{x}} = 27 \]
03

- Calculate the Standard Deviation of the Sample Mean

The standard deviation of the sample mean, \( \sigma_{\bar{x}} \), is given by \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \] Substituting the given values: \[\begin{aligned} \sigma_{\bar{x}} = \frac{6}{\sqrt{15}} = \frac{6}{3.87} \approx 1.55 \end{aligned} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Tendency
Central tendency describes the average or middle value of a set of data. Common measures of central tendency include the mean, median, and mode.
In this exercise, we are particularly interested in the mean, which is the total sum of all values divided by the number of values.
The population mean (\mu) is a measure of the central tendency for the entire population.
For any sample taken from this population, the mean of the sample mean \(\mu_{\bar{x}}\) is also equal to the population mean. Understanding central tendency helps to summarize a large set of data with a single value that represents the center of the distribution.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In other words, it tells us how spread out the values are around the mean.
In this case, the population standard deviation (\sigma) is given as 6. To find the standard deviation of the sample mean \(\sigma_{\bar{x}}\), we use the formula:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \].
The standard deviation of the sample mean shows how much the sample mean deviates from the population mean. Smaller sample standard deviations indicate that the sample means are closely clustered around the population mean, while larger standard deviations indicate more spread.
Sample Size
Sample size (\(n\)) is the number of observations in a sample. It plays a crucial role in statistical calculations and directly affects the standard deviation of the sample mean.
In this exercise, the sample size is provided as 15. When calculating the standard deviation of the sample mean, the formula involves the square root of the sample size:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \].
This means that as the sample size increases, the standard deviation of the sample mean decreases, indicating more precision. Conversely, smaller sample sizes lead to higher variability in the sample mean.
Population Mean
The population mean (\(\mu\)) is the average of all values in a population, and it represents the central value of the entire dataset.
The formula for the population mean is:
\[ \mu = \frac{\sum_{i=1}^{N} X_{i}}{N} \],
where \(\sum\_{i=1}^{N} X\_{i}\)) is the sum of all values in the population, and \(\N\) is the total number of values.
In this exercise, the population mean is given as 27. This value helps in understanding the central tendency of the population. When calculating the sample mean, we see that the mean of the sample mean \(\mu_\bar\{x\}\) equals the population mean, showing that the sample's average accurately represents the population.

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Most popular questions from this chapter

The length of human pregnancies is approximately normally distributed with mean \(\mu=266\) days and standard deviation \(\sigma=16\) days. (a) What is the probability a randomly selected pregnancy lasts less than 260 days? (b) Suppose a random sample of 20 pregnancies is obtained. Describe the sampling distribution of the sample mean length of human pregnancies. (c) What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less? (d) What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less? (e) What might you conclude if a random sample of 50 pregnancies resulted in a mean gestation period of 260 days or less? (f) What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Determine \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) from the given parameters of the population and the sample size. \(\mu=80, \sigma=14, n=49\)

The upper leg length of 20 - to 29 -year-old males is normally distributed with a mean length of \(43.7 \mathrm{~cm}\) and a standard deviation of \(4.2 \mathrm{~cm} .\) Source: "Anthropometric Reference Data for Children and Adults: U.S. Population, 1999-2002"; Volume 361, July 7, 2005. (a) What is the probability that a randomly selected 20 - to 29 -yearold male has an upper leg length that is less than \(40 \mathrm{~cm} ?\) (b) A random sample of 9 males who are 20 to 29 years old is obtained. What is the probability that the mean upper leg length is less than \(40 \mathrm{~cm} ?\) (c) What is the probability that a random sample of 12 males who are \(20-29\) years old results in a mean upper leg length that is less than \(40 \mathrm{~cm} ?\) (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. (e) A random sample of 15 males who are \(20-29\) years old results in a mean upper leg length of \(46 \mathrm{~cm} .\) Do you find this result unusual? Why?

The reading speed of second grade students is approximately normal, with a mean of 90 words per minute (wpm) and a standard deviation of 10 wpm. (a) What is the probability a randomly selected student will read more than 95 words per minute? (b) What is the probability that a random sample of 12 second grade students results in a mean reading rate of more than 95 words per minute? (c) What is the probability that a random sample of 24 second grade students results in a mean reading rate of more than 95 words per minute? (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. (e) A teacher instituted a new reading program at school. After 10 weeks in the program, it was found that the mean reading speed of a random sample of 20 second grade students was 92.8 wpm. What might you conclude based on this result? (f) There is a \(5 \%\) chance that the mean reading speed of a random sample of 20 second grade students will exceed what value?

The quality-control manager of a Long John Silver's restaurant wants to analyze the length of time that a car spends at the drive-through window waiting for an order. It is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1 seconds. The distribution of time at the window is skewed right (data based on information provided by Danica Williams, student at Joliet Junior College). (a) To obtain probabilities regarding a sample mean using the normal model, what size sample is required? (b) The quality-control manager wishes to use a new delivery system designed to get cars through the drive-through system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. What is the probability of obtaining a sample mean of 56.8 seconds or less, assuming that the population mean is 59.3 seconds? Do you think that the new system is effective? (c) Treat the next 50 cars that arrive as a simple random sample. There is a \(15 \%\) chance the mean time will be at or below ____ seconds
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