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Determine \(\mu_{x}\) and \(\sigma_{x}\) from the given parameters of the population and the sample size. $$\mu=52, \sigma=10, n=21$$

Short Answer

Expert verified
Sample mean \( \mu_{x} = 52 \) and standard error \( \sigma_{x} \) is approximately 2.18.

Step by step solution

01

Identify Given Parameters

The problem provides the population mean \( \mu = 52 \), the population standard deviation \( \sigma = 10 \), and the sample size \( \ = 21 \). These values will be used to find the sample mean and standard deviation.
02

Understand Population and Sample Mean

For a random sample from a population, the sample mean \( \mu_{x} \) is equal to the population mean \( \mu \). Therefore, \( \mu_{x} = 52 \).
03

Calculate Standard Error

The standard error of the sample mean (\( \sigma_{x} \)) is given by \( \sigma_{x} = \frac{\sigma}{\sqrt{n}} \). Substitute the given values to find it: \[\sigma_{x} = \frac{10}{\sqrt{21}} \approx \frac{10}{4.5826} \approx 2.18 \]
04

Summarize Results

The sample mean \( \mu_{x} \) is equal to 52 and the standard error \( \sigma_{x} \) is approximately 2.18.

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Key Concepts

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

Population Mean
When we talk about the population mean, we refer to the average value of a particular characteristic within an entire population. The population mean, represented by the symbol \( \mu \), is a central concept in statistics. It offers a measure of the central tendency of the data.

To calculate the population mean, you sum all the values in the population and then divide by the number of values.

For instance, in our exercise, the population mean is provided directly as \( \mu = 52 \). This means that, for the entire population, the average value is 52. Hence, the population mean shows us this common value found within the population.
Sample Mean
The sample mean is a very important statistical measure when working with data from a sample instead of an entire population.
The sample mean, denoted \( \mu_{x} \), is basically the average value of a characteristic within a sample set. This gives us an estimation of the population mean.

Interestingly, when a sample is randomly selected from the population, the sample mean \( \mu_{x} \) is equal to the population mean \( \mu \). From our exercise:
  • Population Mean \( \mu=52 \)
  • Sample Mean \( \mu_{x}=52 \)

So, here we see that \( \mu_{x} = 52 \) as well. This is because our sample mean is intended to reflect the characteristics of the population.
Standard Error
The standard error helps us understand how much the sample mean \( \mu_{x} \) is expected to vary from the population mean \( \mu \). It's denoted as \( \sigma_{x} \). By calculating the standard error, we can gauge the accuracy of our sample mean as an estimate of the population mean.

The standard error is calculated using the formula:
\[ \sigma_{x} = \frac{\sigma}{\sqrt{n}} \]

where:
  • \( \sigma \) is the population standard deviation (10 in this exercise).
  • \( \sqrt{n} \) is the square root of the sample size (here, \( \sqrt{21} \)).
Plugging in our values:
\[ \sigma_{x} = \frac{10}{\sqrt{21}} \]
\[........ = \frac{10}{4.5826} \]\[........ \approx 2.18 \]
Thus, the standard error \( \sigma_{x} \) is approximately 2.18. This gives us an idea about the variability of sample means if we were taking multiple samples from the same population.

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Most popular questions from this chapter

The most famous geyser in the world, Old Faithful in Yellowstone National Park, has a mean time between eruptions of 85 minutes. If the interval of time between eruptions is normally distributed with standard deviation 21.25 minutes, answer the following questions: (Source: www.unmuseum.org) (a) What is the probability that a randomly selected time interval between eruptions is longer than 95 minutes? (b) What is the probability that a random sample of 20 time intervals between eruptions has a mean longer than 95 minutes? (c) What is the probability that a random sample of 30 time intervals between eruptions has a mean longer than 95 minutes? (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. (e) What might you conclude if a random sample of 30 time intervals between eruptions has a mean longer than 95 minutes?

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

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?

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\&P 500 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.

Suppose a simple random sample of size \(n=200\) is obtained from a population whose size is \(N=25,000\) and whose population proportion with a specified characteristic is \(p=0.65\) (a) Describe the sampling distribution of \(\hat{p}\) (b) What is the probability of obtaining \(x=136\) or more individuals with the characteristic? That is, what is \(P(\hat{p} \geq 0.68) ?\) (c) What is the probability of obtaining \(x=118\) or fewer individuals with the characteristic? That is, what is \(P(\hat{p} \leq 0.59) ?\)

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