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Chapter 7: Problem 82

The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited. Would you be surprised, based upon numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes? a. yes b. no c. There is not enough information.

Short Answer

Expert verified
a. yes

Step by step solution

01

Understand the Uniform Distribution

The time to wait for a bus is uniformly distributed from 0 to 75 minutes. This means each minute from 0 to 75 is equally likely. The mean of a uniform distribution is calculated as \( \mu = \frac{a+b}{2} \), where \( a \) is the minimum value and \( b \) is the maximum value.
02

Calculate the Mean of the Distribution

The range of the wait time is from 0 to 75 minutes. Using the formula for the mean of a uniform distribution: \( \mu = \frac{0 + 75}{2} = 37.5 \) minutes. This is the expected average wait time for the distribution.
03

Calculate the Standard Deviation of the Uniform Distribution

The formula for the standard deviation for a uniform distribution is \( \sigma = \frac{b-a}{\sqrt{12}} \). Substituting the given values: \( \sigma = \frac{75-0}{\sqrt{12}} \approx 21.65 \) minutes.
04

Determine the Standard Error of the Sample Mean

The standard error (SE) of the sample mean is calculated as \( SE = \frac{\sigma}{\sqrt{n}} \), where \( n = 100 \). Thus, \( SE = \frac{21.65}{\sqrt{100}} = 2.165 \) minutes.
05

Calculate the Z-Score for the Sample Mean

To find how surprising the sample mean of 30 minutes is, calculate the z-score: \( z = \frac{\bar{x} - \mu}{SE} \), where \( \bar{x} = 30 \) and \( \mu = 37.5 \). Thus, \( z = \frac{30 - 37.5}{2.165} \approx -3.46 \).
06

Conclude Based on the Z-Score

A z-score of -3.46 is quite low, indicating that a sample mean of 30 minutes is more than 3 standard deviations below the expected mean of 37.5 minutes. This corresponds to a very low probability event (much less than 5%). Therefore, it would be surprising if the average wait time for 100 riders was less than 30 minutes.

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Key Concepts

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

Mean Calculation
A mean is the average of a set of numbers and provides a central value for the data set. For a uniform distribution, which in this case is the wait time for a bus duration equally probable between 0 and 75 minutes, the mean is determined using the formula: \[ \mu = \frac{a+b}{2} \] where \( a \) is the minimum wait time and \( b \) is the maximum wait time.
  • Given \( a = 0 \) minutes and \( b = 75 \) minutes, substituting into the formula yields:
  • \( \mu = \frac{0 + 75}{2} = 37.5 \) minutes.
This value indicates that on average, one would expect to wait 37.5 minutes for a bus. Calculating the mean provides an important reference, allowing us to compare actual data against expected outcomes.
Standard Deviation
The concept of standard deviation deals with how spread out the numbers are from the mean in a data set. For a uniform distribution, the standard deviation gives an idea of how much variability exists in the wait times. The formula specific to the standard deviation of a uniform distribution is: \[ \sigma = \frac{b-a}{\sqrt{12}} \] Using the bus wait times:
  • \( a = 0 \) and \( b = 75 \)
  • Plugging into the formula, we find: \( \sigma = \frac{75-0}{\sqrt{12}} \approx 21.65 \) minutes.
A higher standard deviation indicates greater diversity in bus wait times, while a lower standard deviation would suggest they are more uniform.
Standard Error
Standard error is crucial in inferential statistics, as it provides insight into the precision of a sample mean in estimating the population mean. It's computed from the standard deviation and the sample size, \( n \). The formula is: \[ SE = \frac{\sigma}{\sqrt{n}} \]
  • Here, \( \sigma = 21.65 \) minutes and \( n = 100 \).
  • Calculating yields, \( SE = \frac{21.65}{\sqrt{100}} = 2.165 \) minutes.
This value tells us how much the sample mean of 100 riders' bus wait times might vary from the true mean (37.5 minutes). The smaller the standard error, the closer your sample mean is expected to be to the true mean.
Z-Score
The Z-score tells us how many standard deviations an element is from the mean. It helps determine whether a sample mean, like the average wait time for 100 riders, is typical or unusual. The formula to calculate the z-score is: \[ z = \frac{\bar{x} - \mu}{SE} \] Where:
  • \( \bar{x} = 30 \) minutes, the sample mean.
  • \( \mu = 37.5 \) minutes, the mean of the uniform distribution.
  • \( SE = 2.165 \) minutes.
Calculating gives: \[ z = \frac{30 - 37.5}{2.165} \approx -3.46 \]A Z-score of -3.46 means the sample mean is significantly lower than expected, here more than 3 standard deviations below the mean. This suggests our sample mean is indeed surprising and a rare occurrence under this distribution, aligning closely with statistical benchmarks, where events below a z-score of -2 are considered very uncommon.

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Most popular questions from this chapter

Richard鈥檚 Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery. The average wait time is: a. one hour. b. two hours. c. two and a half hours. d. four hours.

Richard鈥檚 Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least one and a half more hours is: a. \(\frac{1}{4}\) b. \(\frac{1}{2}\) c. \(\frac{3}{4}\) d. \(\frac{3}{8}\)

A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken. Find the \(90^{\text { th }}\) percentile for the mean weight for the 100 weights.

Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children. a. In words, \(X=\) ______ b. The distribution is ____. c. In words, \(\Sigma X=\) ______ d. \(\Sigma X \sim\) ____(____,____) e. Find the probability that the total weight of open boxes is less than 250 pounds. f. Find the 35th percentile for the total weight of open boxes of cereal.

An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = the object of interest. What is the sample size if the mean of \(\Sigma X\) is \(15,200 ?\)
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