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The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, \(\mu=0.5\) and \(\sigma=0.289\) a. If \(\bar{x}\) is the average waiting time for a random sample of \(n=16\) waiting times, what are the values of the mean and standard deviation of the sampling distribution of \(\bar{x} ?\) b. Answer Part (a) for a random sample of 50 waiting times. Draw a picture of the approximate sampling distribution of \(\bar{x}\) when \(n=50\).

Short Answer

Expert verified
a. For a random sample of 16 waiting times, the mean of the sampling distribution of \(\bar{x}\) is \(\mu_{\bar{x}} = 0.5\) and the standard deviation is \(\sigma_{\bar{x}} = 0.07225\). b. For a random sample of 50 waiting times, the mean of the sampling distribution of \(\bar{x}\) is \(\mu_{\bar{x}} = 0.5\) and the standard deviation is \(\sigma_{\bar{x}} \approx 0.0409\). The sampling distribution of \(\bar{x}\) when \(n=50\) will be approximately a normal distribution, centered at 0.5 and with a bell-shaped curve.

Step by step solution

01

Mean of the sampling distribution of \(\bar{x}\) for \(n=16\)#

The mean of the sampling distribution of \(\bar{x}\) for \(n=16\) is equal to the population mean, which is given as \(\mu=0.5\). So, $$\mu_{\bar{x}} = \mu = 0.5$$.
02

Standard deviation of the sampling distribution of \(\bar{x}\) for \(n=16\)#

To calculate the standard deviation of the sampling distribution of \(\bar{x}\) for \(n=16\), we can use the formula: $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$$Plug in the values: $$\sigma_{\bar{x}} = \frac{0.289}{\sqrt{16}} = \frac{0.289}{4} = 0.07225$$ b. Calculate the mean and standard deviation for a random sample of 50 waiting times.
03

Mean of the sampling distribution of \(\bar{x}\) for \(n=50\)#

The mean of the sampling distribution of \(\bar{x}\) for \(n=50\) is equal to the population mean, which is given as \(\mu=0.5\). So, $$\mu_{\bar{x}} = \mu = 0.5$$.
04

Standard deviation of the sampling distribution of \(\bar{x}\) for \(n=50\)#

To calculate the standard deviation of the sampling distribution of \(\bar{x}\) for \(n=50\), we can use the formula: $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$$Plug in the values: $$\sigma_{\bar{x}} = \frac{0.289}{\sqrt{50}} \approx 0.0409$$
05

Sampling distribution of \(\bar{x}\) when \(n=50\)#

When \(n=50\), the sampling distribution of \(\bar{x}\) will be approximately a normal distribution with a mean (\(\mu_{\bar{x}}\)) of 0.5 and a standard deviation (\(\sigma_{\bar{x}}\)) of 0.0409. The shape of the distribution will be bell-shaped, centered at the mean, and with the standard deviation defining the spread of the distribution.

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Key Concepts

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

Uniform Distribution
The uniform distribution is a type of probability distribution where all outcomes are equally likely. Think of it like a perfectly flat line where each point on the line has the same chance of being chosen. In the context of waiting times for an elevator, a uniform distribution from 0 to 1 minute means that it's just as likely for someone to wait 10 seconds as it is to wait 50 seconds.

Every part of the interval between 0 and 1 minute is 'uniform,' and there's no concentration of likelihood in any particular region. This distribution is characterized by two parameters: the minimum value and the maximum value, signaling the bounds within which each observation is equally likely to fall. The waiting time example provided gives a clear real-world scenario of how uniform distribution works, making it easier to visualize and grasp the abstract concept.
Standard Deviation
Standard deviation is a measure of how spread out numbers are in a data set. Imagine you and your friends are throwing darts. If everyone's darts are close together on the dartboard, the spread or standard deviation is low. But if everyone's darts are scattered all over the board, the standard deviation is high.

In the elevator scenario, standard deviation tells us how much variation there is from the average (mean) waiting time. If the standard deviation is small (like in the case when n=50, and the standard deviation of the sample mean is approximately 0.0409), it indicates that most people's waiting time is pretty close to the average 30 seconds. Equations, like the one used in the solution (\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\]), allow us to calculate this standard deviation for a set of samples, providing critical insight into the variability of the waiting times.
Central Limit Theorem
The Central Limit Theorem (CLT) is like the secret sauce that allows us to use normal distribution methods on all sorts of data, even if the data isn't normally distributed. It says that if you take a large enough sample size from any distribution (like the uniform distribution of our elevator waiting times), the sampling distribution of the sample mean will be approximately normally distributed.

This theorem is vital because it means that we can make predictions about the sample means using properties of the normal distribution. For example, as the problem illustrates, the sample mean of the waiting times will form a bell-shaped curve centered at the population mean (\(\mu = 0.5\)) as the sample size grows larger, in this case to 50. The spread of this distribution, which we calculated using the standard deviation formula, tells us how much the average waiting time of different samples will vary, allowing for all sorts of statistical magic, like confidence intervals and hypothesis testing.

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Most popular questions from this chapter

Suppose that college students with a checking account typically write relatively few checks in any given month, whereas people who are not college students typically write many more checks during a month. Suppose that \(50 \%\) of a bank's accounts are held by students and that \(50 \%\) are held by people who are not college students. Let \(x\) represent the number of checks written in a given month by a randomly selected bank customer. a. Give a sketch of what the population distribution of \(x\) might look like. b. Suppose that the mean value of \(x\) is 22.0 and that the standard deviation is \(16.5 .\) If a random sample of \(n=\) 100 customers is to be selected and \(\bar{x}\) denotes the sample mean number of checks written during a particular month, where is the sampling distribution of \(\bar{x}\) centered, and what is the standard deviation of the sampling distribution of \(\bar{x}\) ? Sketch a rough picture of the sampling distribution. c. What is the approximate probability that \(\bar{x}\) is at most \(20 ?\) At least \(25 ?\)

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(n=16, t=1.6\) b. Upper-tailed test, \(n=14, t=3.2\) c. Lower-tailed test, \(n=20, t=-5.1\) d. Two-tailed test, \(n=16, t=6.3\)

The paper "The Effects of Adolescent Volunteer Activities on the Perception of Local Society and Community Spirit Mediated by Self-Conception" (Advanced Science and Technology Letters [2016]: 19-23) describes a survey of a large representative sample of middle school children in South Korea. One question in the survey asked how much time per year the children spent in volunteer activities. The sample mean was 14.76 hours and the sample standard deviation was 16.54 hours. a. Based on the reported sample mean and sample standard deviation, explain why it is not reasonable to think that the distribution of volunteer times for the population of South Korean middle school students is approximately normal. b. The sample size was not given in the paper, but the sample size was described as "large." Suppose that the sample size was 500 . Explain why it is reasonable to use a one-sample \(t\) confidence interval to estimate the population mean even though the population distribution is not approximately normal. c. Calculate and interpret a confidence interval for the mean number of hours spent in volunteer activities per year for South Korean middle school children. (Hint: See Example 12.7.)

Suppose that the population mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a random sample of \(n=25\) adult males is to be selected, what is the probability that the sample mean distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? At least \(68 \mathrm{~mm}\) ? b. Suppose that a random sample of 100 adult males is to be selected. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 \(\mathrm{mm}\) ? At least \(68 \mathrm{~mm} ?\)

An automobile manufacturer decides to carry out a fuel efficiency test to determine if it can advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon). Six people each drive a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{ll}30.3 & 29.6\end{array}\) \(\begin{array}{llll}27.2 & 29.3 & 31.2 & 28.4\end{array}\) Assuming that fuel efficiency is normally distributed, do these data provide evidence against the claim that actual mean fuel efficiency for this model is (at least) \(30 \mathrm{mpg}\) ?

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