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The time that a randomly selected individual waits for an elevator in an office building has a uniform distribution over the interval from 0 to \(1 \mathrm{~min}\). It can be shown that for this distribution \(\mu=0.5\) and \(\sigma=0.289\). a. Let \(\bar{x}\) be the sample average waiting time for a random sample of 16 individuals. What are the mean and standard deviation of the sampling distribution of \(\bar{x}\) ? b. Answer Part (a) for a random sample of 50 individuals. In this case, sketch a picture of a good approximation to the actual \(\bar{x}\) distribution.

Short Answer

Expert verified
a) For a sample of 16 individuals, the mean is 0.5 min and the standard deviation is 0.07225 min. b) For a sample of 50 individuals, the mean is 0.5 min and the standard deviation is 0.04086 min. The sampling distribution approximates a normal distribution centered at 0.5 min with a standard deviation of 0.04086 min.

Step by step solution

01

Find the Mean of the Sampling Distribution

The mean of the sampling distribution of the sample mean is always equal to the mean of the population. Thus, for both samples of 16 and 50 individuals, the mean of the sampling distribution of \(\bar{x}\) will be the same as the population mean, \(\mu = 0.5\) min.
02

Find the Standard Deviation of the Sampling Distribution for a Sample of 16 Individuals

The standard deviation of the sampling distribution of \(\bar{x}\) is equal to the population standard deviation \(\sigma\) divided by the square root of the sample size \(n\). Therefore, for a sample of 16 individuals, the standard deviation of the sampling distribution of \(\bar{x}\) is \(\sigma / \sqrt{n} = 0.289 / \sqrt{16} = 0.07225\) min.
03

Find the Standard Deviation of the Sampling Distribution for a Sample of 50 Individuals

Again use the formula \(\sigma / \sqrt{n}\), but this time with a sample size of 50. The standard deviation of the sampling distribution of \(\bar{x}\) for a sample of 50 individuals is \(\sigma / \sqrt{n} = 0.289 / \sqrt{50} = 0.04086\) min.
04

Sketch the Sampling Distribution for a Sample of 50 Individuals

The sampling distribution for a sample of 50 individuals approximates a normal distribution because the sample size is large. The distribution is centered around the mean of 0.5 min with a standard deviation of 0.04086 min. The graph should be bell-shaped.

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Key Concepts

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

Uniform Distribution
The idea of uniform distribution refers to a type of probability distribution where all outcomes are equally likely. Imagine a spinner divided equally into sections - each section represents an outcome; when spun, each outcome has the same chance of occurring.
In the exercise, the waiting time for an elevator follows a uniform distribution between 0 and 1 minute. This means every time point within this interval is equally likely to be the waiting time. The mean, or average waiting time, is calculated as the midpoint of this interval, which is 0.5 minutes.
If you think about it, uniform distribution is like having a perfectly level playing field where everyone gets an equal shot at each outcome. This intuitive concept helps set the stage for understanding more complex statistical ideas.
Central Limit Theorem
The Central Limit Theorem (CLT) is a foundational concept in statistics. It tells us that when you take large enough random samples from a population, the distribution of the sample means will tend to resemble a normal distribution, regardless of the original population distribution.
This is crucial in the exercise where we look at the sample sizes of 16 and 50 individuals. When the sample size is 50, the sampling distribution of the sample mean begins to look more like a bell curve. This is thanks to the CLT, enabling us to use statistical methods that require normality.
The beauty of the CLT lies in its ability to connect individual random variables to the predictable nature of the sample mean's distribution, allowing statisticians to make probabilistic statements and predictions.
Standard Deviation
Standard deviation, often represented by the symbol \(\sigma\), is a measure of how spread out the values in a data set are. It gives us an idea of how much the values differ from the mean. In the context of the uniform distribution mentioned earlier, it was given as 0.289 minutes.
A key part of the exercise was finding the standard deviation of the sampling distribution—often called the standard error of the mean—when the sample sizes are 16 and 50. This is calculated by the formula: \[ \sigma / \sqrt{n} \], where \(\sqrt{n}\), is the square root of the sample size. This formula shows as your sample size increases, the standard error decreases, indicating more precise estimates of the population mean.
Understanding how standard deviation applies to both the population and the sampling distribution is critical for statistical analysis.
Sample Size
Sample size, denoted by \(n\), plays a vital role in statistics. It refers to the number of observations or data points collected in a study. In the exercise, samples of 16 and 50 individuals were considered.
A larger sample size tends to give a more accurate estimate of the population parameters, like the mean. The reason is that larger samples reduce the standard error, the variability of the sample mean, making the sample mean a better estimate of the population mean.
Moreover, as seen in the exercise, a larger sample size, which was 50, meant that the sampling distribution was more aligned with a normal distribution due to the Central Limit Theorem. In essence, ensuring an adequately large sample size is key to obtaining reliable and valid statistical conclusions.

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Most popular questions from this chapter

An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable \(x\) with a mean value of \(50 \mathrm{lb}\) and a standard deviation of \(20 \mathrm{lb}\). If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With \(n=100\), the total weight exceeds the limit when the average weight \(\bar{x}\) exceeds \(6000 / 100\).)

Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=2 \mathrm{~min}\) and \(\sigma=0.8\) min, respectively. a. If \(\bar{x}\) is the sample average time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?

Suppose that \(20 \%\) of the subscribers of a cable television company watch the shopping channel at least once a week. The cable company is trying to decide whether to replace this channel with a new local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep the shopping channel if the sample proportion is greater than \(.25 .\) What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only \(.20 ?\)

The article "Thrillers" (Newsweek, April 22,1985 ) stated, "Surveys tell us that more than half of America's college graduates are avid readers of mystery novels." Let \(\pi\) denote the actual proportion of college graduates who are avid readers of mystery novels. Consider a sample proportion \(p\) that is based on a random sample of 225 college graduates. a. If \(\pi=.5\), what are the mean value and standard deviation of \(p ?\) Answer this question for \(\pi=.6\). Does \(p\) have approximately a normal distribution in both cases? Explain. b. Calculate \(P(p \geq .6)\) for both \(\pi=.5\) and \(\pi=.6\). c. Without doing any calculations, how do you think the probabilities in Part (b) would change if \(n\) were 400 rather than \(225 ?\)

What is the difference between \(\bar{x}\) and \(\mu\) ? between \(s\) and \(\sigma\) ?

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