/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Determine \(\mu_{\bar{x}}\) and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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: \[\begin{aligned} \sigma_{\bar{x}} = \frac{6}{\sqrt{15}} = \frac{6}{3.87} \approx 1.55 \end{aligned} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain what a sampling distribution is.

True or False: The population proportion and sample proportion always have the same value.

A simple random sample of size \(n=40\) is obtained from a population with \(\mu=50\) and \(\sigma=4 .\) Does the population need to be normally distributed for the sampling distribution of \(\bar{x}\) to be approximately normally distributed? Why? What is the sampling distribution of \(\bar{x} ?\)

The S\&P 500 is a collection of 500 stocks of publicly traded companies. Using data obtained from Yahoo! Finance, the monthly rates of return of the S\&P500 since 1950 are normally distributed. The mean rate of return is \(0.007233(0.7233 \%),\) and the standard deviation for rate of return is \(0.04135(4.135 \%)\). (a) What is the probability that a randomly selected month has a positive rate of return? That is, what is \(P(x>0) ?\) (b) Treating the next 12 months as a simple random sample, what is the probability that the mean monthly rate of return will be positive? That is, with \(n=12,\) what is \(P(\bar{x}>0) ?\) (c) Treating the next 24 months as a simple random sample, what is the probability that the mean monthly rate of return will be positive? (d) Treating the next 36 months as a simple random sample, what is the probability that the mean monthly rate of return will be positive? (e) Use the results of parts (b)-(d) to describe the likelihood of earning a positive rate of return on stocks as the investment time horizon increases.

The shape of the distribution of the time required to get an oil change at a 10 -minute oil-change facility is unknown. However, records indicate that the mean time for an oil change is 11.4 minutes, and the standard deviation for oilchange time is 3.2 minutes. (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? (b) What is the probability that a random sample of \(n=40\) oil changes results in a sample mean time of less than 10 minutes? (c) Suppose the manager agrees to pay each employee a \(\$ 50\) bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, what mean oil-change time would there be a \(10 \%\) chance of being at or below? This will be the goal established by the manager.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.