/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 What is the age distribution of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the age distribution of adult shoplifters ( 21 years of age or older) in supermarkets? The following is based on information taken from the National Retail Federation. A random sample of 895 incidents of shoplifting gave the following age distribution: \begin{tabular}{l|ccc} \hline Age range (years) & \(21-30\) & \(31-40\) & 41 and over \\ \hline Number of shoplifters & 260 & 348 & 287 \\ \hline \end{tabular} Estimate the mean age, sample variance, and sample standard deviation for the shoplifters. For the class 41 and over, use \(45.5\) as the class midpoint.

Short Answer

Expert verified
The mean age is approximately 35.82, the sample variance is 60.82, and the sample standard deviation is 7.80.

Step by step solution

01

Calculate Class Midpoints

First, determine the class midpoint for each age range. For the age range \(21-30\), the midpoint is \(25.5\), calculated as \(\frac{21+30}{2}\). For \(31-40\), the midpoint is \(35.5\) and for \(41\) and over, the given midpoint is \(45.5\). These midpoints represent the typical age for each class.
02

Calculate Weighted Mean Age

To estimate the mean age of the shoplifters, multiply each class midpoint by the number of shoplifters in that age range and sum these products. Then, divide by the total number of shoplifters. The formula is: \( \bar{x} = \frac{\sum (f_i \cdot m_i)}{\sum f_i} \), where \(f_i\) is the frequency and \(m_i\) is the midpoint.\[\bar{x} = \frac{(260 \times 25.5) + (348 \times 35.5) + (287 \times 45.5)}{895} = \frac{6630 + 12354 + 13078.5}{895} = \frac{32062.5}{895} \approx 35.82 \]
03

Calculate Sample Variance

Sample variance is calculated using the formula: \( s^2 = \frac{\sum f_i(m_i - \bar{x})^2}{n-1} \). First, calculate \((m_i - \bar{x})^2\) for each class.- For \(21-30\), \((25.5 - 35.82)^2 = 106.0324\)- For \(31-40\), \((35.5 - 35.82)^2 = 0.1024\)- For \(41+\), \((45.5 - 35.82)^2 = 93.3124\)Now, compute \(\sum f_i(m_i - \bar{x})^2\):\[\sum f_i(m_i - \bar{x})^2 = 260 \times 106.0324 + 348 \times 0.1024 + 287 \times 93.3124 = 27568.424 + 35.6352 + 26773.2688 = 54377.328\]Finally, divide by \(n-1\):\[s^2 = \frac{54377.328}{894} \approx 60.82\]
04

Calculate Sample Standard Deviation

The sample standard deviation \(s\) is the square root of the sample variance \(s^2\). Thus,\[s = \sqrt{60.82} \approx 7.80\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Mean
The sample mean is a fundamental concept in statistics, representing the average value of a sample. It's particularly useful when you need to summarize a large set of data with a single value. In our exercise, we are interested in the mean age of adult shoplifters. To find the sample mean of any set of values, you sum up all the values and divide by the total number of observations.

For group data like age ranges, we use midpoints to estimate the mean. Here, the class midpoints for age ranges provide an estimate of the typical age in each category. For instance, the midpoint for the range 21-30 is 25.5, calculated as \(\frac{21+30}{2}\). Similarly, we compute midpoints for other age ranges and use them in the weighted mean formula:
  • Multiply each midpoint by its corresponding frequency (number of individuals).
  • Add the results to get a total sum.
  • Divide by the total frequency (the sum of all frequencies), which in this case is 895.
Hence, the sample mean age is approximately 35.82 years, meaning on average, adult shoplifters in this sample tend to be around 36 years old.
Sample Variance
Sample variance measures how much the ages of shoplifters vary around the sample mean. It's an essential statistic because it helps us understand the spread or dispersion of data in a dataset.

To calculate sample variance, follow these steps:
  • First, find the difference between each class midpoint and the calculated sample mean (35.82 years in our case).
  • Square these differences to eliminate negative values and give more weight to outliers.
  • Multiply each squared difference by the frequency of its corresponding class.
  • Sum over all these values to get the "Sum of Squares of Deviations" from the mean.
  • Finally, divide by \(n-1\) to account for degrees of freedom (n is the total number of observations).
The calculations yield a sample variance of approximately 60.82. This value suggests how much the age of shoplifters in this sample differs from the mean age.
Age Distribution
Age distribution refers to how ages are spread across different categories in a dataset. Understanding age distribution helps identify patterns, trends and potential causes or effects associated with different age groups in studies or reports.

In this exercise, age distribution covers three age ranges: 21-30, 31-40, and 41 and over. By looking at the given data:
  • 260 shoplifters are in the 21-30 range,
  • 348 in the 31-40 range,
  • and 287 are 41 and over.
This spread shows which age groups might account for a higher number of shoplifting incidents. It also informs us about the age group with the most frequent occurrences, which here is the 31-40 age group. Exploring age distributions like this can guide policies, prevention strategies, and further studies.
Standard Deviation
Standard deviation is a crucial statistical tool that quantifies the amount of variance or dispersion in a set of data points. It tells us how much the individual data points differ from the mean of the dataset.

The standard deviation is derived from the variance by taking the square root of the sample variance. This step brings the measure back to the original unit of measure (in this case, years), making it more directly interpretable.

For the shoplifting age data, with a sample variance of 60.82, the standard deviation is approximately \(s = \sqrt{60.82} \approx 7.80\).

This indicates that the ages of the shoplifters typically vary by about 7.8 years from the average age of 35.82. In simpler terms, most shoplifters' ages are within 7.8 years above or below the mean age, providing a clear snapshot of age variability in this sample. Standard deviation helps paint a clear picture of how concentrated or spread the age data is around the mean.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One standard for admission to Redfield College is that the student must rank in the upper quartile of his or her graduating high school class. What is the minimal percentile rank of a successful applicant?

Leisure: Maui Vacation How expensive is Maui? If you want a vacation rental condominium (up to four people), visit the Brase/Brase statistics site, find the link to Maui, and then search for accommodations. The Maui News gave the following costs in dollars per day for a random sample of condominiums located throughout the island of Maui. \(\begin{array}{llrlllllll}89 & 50 & 68 & 60 & 375 & 55 & 500 & 71 & 40 & 350 \\ 60 & 50 & 250 & 45 & 45 & 125 & 235 & 65 & 60 & 130\end{array}\) (a) Compute the mean, median, and mode for the data. (b) Compute a \(5 \%\) trimmed mean for the data, and compare it with the mean computed in part (a). Does the trimmed mean more accurately reflect the general level of the daily rental costs? (c) If you were a travel agent and a client asked about the daily cost of renting a condominium on Maui, what average would you use? Explain. Is there any other information about the costs that you think might be useful, such as the spread of the costs?

Consider the mode, median, and mean. Which average represents the middle value of a data distribution? Which average represents the most frequent value of a distribution? Which average takes all the specific values into account?

How hot does it get in Death Valley? The following data are taken from a study conducted by the National Park System, of which Death Valley is a unit. The ground temperatures \(\left({ }^{\circ} \mathrm{F}\right)\) were taken from May to November in the vicinity of Furnace Creek. \(\begin{array}{lllllll}146 & 152 & 168 & 174 & 180 & 178 & 179 \\ 180 & 178 & 178 & 168 & 165 & 152 & 144\end{array}\) Compute the mean, median, and mode for these ground temperatures.

Pax World Balanced is a highly respected, socially responsible mutual fund of stocks and bonds (see Viewpoint). Vanguard Balanced Index is another highly regarded fund that represents the entire U.S. stock and bond market (an index fund). The mean and standard deviation of annualized percent returns are shown below. The annualized mean and standard deviation are based on the years 1993 through 2002 (Source: Morningstar). Pax World Balanced: \(\bar{x}=9.58 \% ; s=14.05 \%\) Vanguard Balanced Index: \(\bar{x}=9.02 \% ; s=12.50 \%\) (a) Compute the coefficient of variation for each fund. If \(\bar{x}\) represents return and \(s\) represents risk, then explain why the coefficient of variation can be taken to represent risk per unit of return. From this point of view, which fund appears to be better? Explain. (b) Compute a \(75 \%\) Chebyshev interval around the mean for each fund. Use the intervals to compare the two funds. As usual, past performance does not guarantee future performance.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.