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Determine \(\mu_{x}\) and \(\sigma_{x}\) from the given parameters of the population and the sample size. $$\mu=64, \sigma=18, n=36$$

Short Answer

Expert verified
\( \text{μ}_x = 64 \) and \( \text{σ}_x = 3 \)

Step by step solution

01

- Understand the Given Parameters

Identify the given information: - Population mean, \(\text{μ} = 64\) - Population standard deviation, \(\text{σ} = 18\) - Sample size, \(n = 36\)
02

- Calculate the Sample Mean \(\text{μ}_x\)

The sample mean \( \text{μ}_x \) of the population is the same as the population mean \( μ \). Thus, \( \text{μ}_x = 64 \).
03

- Calculate the Sample Standard Deviation \(\text{σ}_x\)

Use the formula for the standard deviation of the sampling distribution of the sample mean: \[ \text{σ}_x = \frac{σ}{\text{√n}} \] Substituting the given values, \[ \text{σ}_x = \frac{18}{\text{√36}} = \frac{18}{6} = 3 \]

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Key Concepts

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

Population Mean
The population mean, often represented by the Greek letter \( \mu \), is a measure of the central tendency of an entire population. It is the average of all the values in the population. In the given exercise, the population mean \( \mu \) is 64. This means that when you sum up all the individual values in the population and divide by the number of values, you get 64.

Understanding the population mean is essential because it helps us know the central value around which the data points are distributed. It serves as a benchmark when comparing sample data to the overall population.

In a real-world scenario, the population could be a large set of data like the heights of all students in a school, the annual income of all citizens in a city, or any other comprehensive dataset.
Sample Mean
The sample mean, denoted as \( \mu_x \), is the average of a subset (sample) taken from the population. One important aspect of the sample mean is that it is an unbiased estimator of the population mean. This means that over many samples, the average of the sample means will equal the population mean.

In our exercise, calculating the sample mean is straightforward because it is given that \( \mu = 64 \). Therefore, \( \mu_x \) is also 64. This equality holds because the sample mean is just an average of randomly picked values from a population, and in expectation, it should equal the population mean.

In practical terms, the sample mean allows us to make inferences about the population mean when it's not feasible to collect data from every member of the population. For instance, if you wanted to estimate the average height of students at a school, you might measure a sample of students rather than the entire student body.
Standard Deviation
Standard deviation, represented by the Greek letter \( \sigma \), measures the dispersion or spread of values in a dataset relative to its mean. The population standard deviation provides insight into how much individual data points differ from the mean. In this exercise, the population standard deviation \( \sigma \) is 18.

When dealing with samples, we often calculate the standard deviation of the sampling distribution of the sample mean, denoted as \( \sigma_x \). This is obtained by dividing the population standard deviation by the square root of the sample size: \( \sigma_x = \frac{\sigma} {\sqrt{n}} \)

In our example, given the population standard deviation of 18 and a sample size \( n = 36 \), we find the sample standard deviation \( \sigma_x \) as follows:
\[ \sigma_x = \frac{18}{\sqrt{36}} = \frac{18}{6} = 3. \]

This calculation is crucial because the smaller standard deviation in the sample mean's distribution indicates that sample means tend to be closer to the population mean, enhancing the reliability of statistical inferences.

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Most popular questions from this chapter

A researcher studying ADHD among teenagers obtains a simple random sample of 100 teenagers aged 13 to 17 and asks them whether or not they have ever been prescribed medication for ADHD. To say that the distribution of \(\hat{p},\) the sample proportion who respond no, is approximately normal, how many more teenagers aged 13 to 17 does the researcher need to sample if (a) \(90 \%\) of all teenagers aged 13 to 17 have never been prescribed medication for ADHD? (b) \(95 \%\) of all teenagers aged 13 to 17 have never been prescribed medication for ADHD?

Suppose a simple random sample of size \(n=36\) is obtained from a population with \(\mu=64\) and \(\sigma=18\) (a) Describe the sampling distribution of \(\bar{x}\). (b) What is \(P(\bar{x}<62.6) ?\) (c) What is \(P(\bar{x} \geq 68.7) ?\) (d) What is \(P(59.8<\bar{x}<65.9) ?\)

Burger King's Drive-Through Suppose cars arrive at Burger King's drive-through at the rate of 20 cars every Hour between 12: 00 noon and 1: 00 P.M. A random sample \- of 40 one-hour time periods between 12: 00 noon and 1: 00 P.M. is selected and has 22.1 as the mean number of cars erriving. (a) Why is the sampling distribution of \(\bar{x}\) approximately normal? (b) What is the mean and standard deviation of the sampling distribution of \(\bar{x}\) assuming \(\mu=20\) and \(\boldsymbol{\sigma}=\sqrt{20}\) (c) What is the probability that a simple random sample of 40 one-hour time periods results in a mean of at least 22.1 cars? Is this result unusual? What might we conclude?

True or False: The mean and standard deviation of the distribution of \(\bar{x}\) is \(\mu_{x}=\mu\) and \(\sigma_{x}=\frac{\sigma}{\sqrt{n}},\) respectively, even if the population is not normal.

Based on tests of the Chevrolet Cobalt, engineers have found that the miles per gallon in highway driving are normally distributed, with a mean of 32 miles per gallon and a standard deviation 3.5 miles per gallon. (a) What is the probability that a randomly selected Cobalt gets more than 34 miles per gallon? (b) Suppose that 10 Cobalts are randomly selected and the miles per gallon for each car are recorded. What is the probability that the mean miles per gallon exceed 34 miles per gallon? (c) Suppose that 20 Cobalts are randomly selected and the miles per gallon for each car are recorded. What is the probability that the mean miles per gallon exceed 34 miles per gallon? Would this result be unusual?

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