Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Children going back to school can be expensive for parents-second only to the Christmas holiday season in terms of spending (San Luis Obispo Tribune, August \(18,\) 2005). Parents spend an average of \(\$ 444\) on their children at the beginning of the school year, stocking up on clothes, notebooks, and even iPods. However, not every parent spends the same amount of money. Imagine a data set consisting of the amount spent at the beginning of the school year for each student at a particular elementary school. Would it have a large or a small standard deviation? Explain.

Short Answer

Expert verified
The data set consisting of the amount spent at the beginning of the school year for each student at a particular elementary school would likely have a large standard deviation, indicating a high level of variability in the amounts spent by parents.

Step by step solution

01

Understanding Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values are closer to the average value or the mean, while a high standard deviation implies that the values are spread out over a wider range.
02

Analyzing the Scenario

Given that parents spend an average of \$444 at the beginning of the school year, it indicates that the value is not consistent for every parent. In fact, some parents might spend significantly more or less than this amount. Spending can vary greatly due to a variety of factors such as income level, number of children, need for new items (like clothes, shoes, school supplies), discounts or sales, etc. Thus, expenditure on back-to-school items can vary considerably among different parents.
03

Decoding Standard Deviation for the Given Scenario

Given the variability in spending among parents as discussed in the previous step, it can be concluded that the data set mentioned in the exercise would likely have a high standard deviation. This is because the spending amounts are expected to be spread out over a wide range from the mean of \$444.

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 Vaia!

Key Concepts

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

Statistical Dispersion
When we talk about statistical dispersion, we are referring to the way values in a data set are spread out. It measures how much the numbers in a set differ from the mean, or average, of the quantities. Think of it as a way to describe the diversity or the range of a group of numbers. For instance, if you have a class of students where everyone scored between 90% to 95% on a test, there's low dispersion—everyone performed similarly. On the other hand, if scores ranged from 50% to 100%, the dispersion is high; there's much more variability in the results.

To quantify dispersion, statisticians use different measures, including range, variance, and the interquartile range. However, the most common measure is the standard deviation, which focuses on how spread out the numbers are around the mean score. When the standard deviation is small, it means that the values are clustered closely around the mean; when it's large, the values are more spread out.

In the context of parents' spending on back-to-school supplies, we can expect to see a considerable amount of dispersion. Families have different budgets and shopping habits, which translates into varying levels of expenditure and therefore, a likely higher standard deviation.
Variation in Data Sets
The concept of variation in data sets deals with how individual data points in a set differ from one another. It's an important aspect of statistics as it influences the conclusions we can draw about a population. If we go back to the school spending example, the variation in the amount spent by each parent could result from multiple factors. These include socioeconomic background, the number of children in the family, personal priorities concerning education, and even individual preferences.

It's valuable to explore the sources of variation to understand the dynamics behind a data set better. This could involve considering external factors, like economic conditions and cultural attitudes towards education. Data variation can also be influenced by internal factors, such as personal financial management. By teasing out these factors, we gain a deeper insight into why the data may be showing a high or low variation.

Understanding the variation within a data set is crucial for predictions and decision making. For parents preparing for back-to-school season, appreciating this variation can lead to more tailored budgeting and spending strategies, as well as provide insights for schools and retailers targeting this demographic.
Measuring Spread in Statistics
When we look at a set of numbers, one of the first things we want to understand is how spread out the numbers are. This is what measuring spread in statistics is all about. Among the tools we use for this, the standard deviation is incredibly valuable because it tells us about the concentration of data points around the average value.

Think of the standard deviation like a ruler that helps us understand the range of variation from the 'middle' of our data. It's computed by taking the square root of the average of the squared differences between each data point and the mean. This formula might sound complex, but the essence is that it's a reliable method for getting a feel for how spread out our data is.

Moreover, when the standard deviation is high, it indicates that there's a lot of diversity in the data. This could suggest that there are underlying subgroups or factors contributing to this spread that might need to be investigated further. In the school spending scenario, a high standard deviation would hint at a significant disparity in spending habits, meaning that there's no one-size-fits-all approach when it comes to understanding or predicting parental spending on school supplies.

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

Research by the U.S. Food and Drug Administration (FDA) shows that acrylamide (a possible cancer-causing substance) forms in high-carbohydrate foods cooked at high temperatures, and that acrylamide levels can vary widely even within the same brand of food (Associated Press, December 6,2002 ). FDA scientists analyzed McDonald's french fries purchased at seven different locations and found the following acrylamide levels: \(\begin{array}{lllll}497 & 193 & 328 & 155 & 326\end{array}\) \(\begin{array}{ll}245 & 270\end{array}\) a. Calculate the mean acrylamide level. For each data value, calculate the deviation from the mean. b. Verify that, except for the effect of rounding, the sum of the seven deviations from the mean is equal to 0 forthis data set. (If you rounded the sample mean or the deviations, your sum may not be exactly zero, but it should still be close to zero.) c. Use the deviations from Part (a) to calculate the variance and standard deviation for this data set.

The report "Who Moves? Who Stays Put? Where's Home?" (Pew Social and Demographic Trends, December 17,2008 ) gave the accompanying data on the percentage of the population in a state that was born in the state and is still living there for each of the 50 U.S. states. \(\begin{array}{llllllll}75.8 & 71.4 & 69.6 & 69.0 & 68.6 & 67.5 & 66.7 & 66.3 \\\ 66.1 & 66.0 & 66.0 & 65.1 & 64.4 & 64.3 & 63.8 & 63.7 \\ 62.8 & 62.6 & 61.9 & 61.9 & 61.5 & 61.1 & 59.2 & 59.0 \\ 58.7 & 57.3 & 57.1 & 55.6 & 55.6 & 55.5 & 55.3 & 54.9 \\ 54.7 & 54.5 & 54.0 & 54.0 & 53.9 & 53.5 & 52.8 & 52.5 \\\ 50.2 & 50.2 & 48.9 & 48.7 & 48.6 & 47.1 & 43.4 & 40.4 \\ 35.7 & 28.2 & & & & & & \end{array}\) a. Find the values of the median, the lower quartile, and the upper quartile. b. The two smallest values in the data set are 28.2 (Alaska) and 35.7 (Wyoming). Are these two states outliers? c. Construct a modified boxplot for this data set and comment on the interesting features of the plot.

The San Luis ObispoTelegram-Tribune(October1,1994) reported the following monthly salaries for supervisors from six different counties: \(\$ 5,354\) (Kern), \(\$ 5,166\) (Monterey), \(\$ 4,443\) (Santa Cruz), \(\$ 4,129\) (Santa Barbara), \(\$ 2,500\) (Placer), and \$2,220 (Merced). San Luis Obispo County supervisors are supposed to be paid the average of the two counties in the middle of this salary range. Which measure of center determines this salary, and what is its value? Find the value of the other measure of center featured in this chapter. Why is it not as favorable to the San Luis Obispo County supervisors (although it might appeal to taxpayers)?

The article "Rethink Diversification to Raise Returns, Cut Risk" (San Luis Obispo Tribune, January 21,2006 ) included the following paragraph: In their research, Mulvey and Reilly compared the results of two hypothetical portfolios and used actual data from 1994 to 2004 to see what returns they would achieve. The first portfolio invested in Treasury bonds, domestic stocks, international stocks, and cash. Its 10 -year average annual return was \(9.85 \%\) and its volatility-measured as the standard deviation of annual returns-was \(9.26 \%\). When Mulvey and Reilly shifted some assets in the portfolio to include funds that invest in real estate, commodities, and options, the 10-year return rose to \(10.55 \%\) while the standard deviation fell to \(7.97 \% .\) In short, the more diversified portfolio had a slightly better return and much less risk. Explain why the standard deviation is a reasonable measure of volatility and why a smaller standard deviation means less risk.

The mean number of text messages sent per month by customers of a cell phone service provider is 1,650 , and the standard deviation is \(750 .\) Find the \(z\) -score associated with each of the following numbers of text messages sent. a. 0 b. \(\quad 10,000\) c. \(\quad 4,500\) d. \(\quad 300\)

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.

Sign-up for free