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Chapter 2: Problem 39

A standard deviation contest. This is a standard deviation contest. You must choose four numbers from the whole numbers 0 to 10 , with repeats allowed. (a) Choose four numbers that have the smallest possible standard deviation. (b) Choose four numbers that have the largest possible standard deviation. (c) Is more than one choice possible in either part (a) or (b)? Explain.

Short Answer

Expert verified
(a) Four identical numbers; (b) Numbers like 0, 0, 10, 10; (c) Many choices for smallest deviation, one optimal for largest.

Step by step solution

01

Understanding the Standard Deviation

The standard deviation is a measure of how spread out the numbers in a data set are. A smaller standard deviation means the numbers are closer together, while a larger standard deviation indicates that the numbers are more spread out.
02

Smallest Possible Standard Deviation

To achieve the smallest standard deviation, you need all the numbers to be as close as possible. If they are identical, the variance is zero, which means the standard deviation will also be zero. For instance, choosing the numbers 5, 5, 5, and 5 means there is no deviation.
03

Verification for Smallest Standard Deviation

Let's calculate: the mean of 5, 5, 5, 5 is \[ \mu = \frac{5 + 5 + 5 + 5}{4} = 5 \]The variance \[ \sigma^2 = \frac{(5-5)^2 + (5-5)^2 + (5-5)^2 + (5-5)^2}{4} = 0 \]Thus, the standard deviation is \[ \sigma = \sqrt{0} = 0 \]Hence, choosing any four identical numbers provides the smallest standard deviation.
04

Largest Possible Standard Deviation

For the largest standard deviation, the numbers need to be as spread out as possible. Choose the extreme values, such as 0 and 10, twice each. For instance, the numbers 0, 0, 10, and 10.
05

Verification for Largest Standard Deviation

Check the spread of 0, 0, 10, and 10. Calculate the mean:\[ \mu = \frac{0 + 0 + 10 + 10}{4} = 5 \]Calculate the variance:\[ \sigma^2 = \frac{(0-5)^2 + (0-5)^2 + (10-5)^2 + (10-5)^2}{4} = \frac{25 + 25 + 25 + 25}{4} = 25 \]Thus, the standard deviation is \[ \sigma = \sqrt{25} = 5 \]This configuration provides the largest spread.
06

Analyzing Multiple Choices

For the smallest standard deviation, any four identical integers from 0 to 10 will work. For the largest standard deviation, another option could be alternating values that span the range, but 0, 0, 10, 10 is optimal. There are multiple choices for the smallest deviation but typically one for the largest in this context.

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Key Concepts

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

Variance
Variance is a fundamental concept in statistics that helps us understand how data values are dispersed around the mean. In simple terms, variance tells us how much the numbers in a dataset differ from each other.
It is calculated by finding the average of the squared differences from the mean.
  • To compute variance, we first find the mean (average) of all numbers.
  • Then, we subtract the mean from each number to find the deviation of each number from the mean.
  • We square each of these deviations to eliminate negative values.
  • Finally, we calculate the average of these squared deviations, and this result is the variance.
Variance (\[\sigma^2\]) is important because it provides insight into the spread of the data. The larger the variance, the more spread out the data points are from the mean. If the variance is zero, all data points are identical and equal to the mean.
Mean Calculation
The mean, often referred to as the average, is the central value of a dataset. It is calculated by adding together all the numbers and then dividing by the number of values in the dataset.
This measure provides us with a simple, one-number summary of the data.For instance, if we consider the numbers 5, 5, 5, and 5:
  • Add all numbers together: 5 + 5 + 5 + 5 = 20.
  • Divide the sum by the number of numbers (4 in this case): \[ \mu = \frac{20}{4} = 5 \]
The mean is crucial because it represents the balance point of the dataset, and it is the basis for calculating both variance and standard deviation. When comparing datasets, the mean provides a quick way to determine central tendency and to perform other statistical analyses.
Data Spread
Data spread refers to how much the numbers in a dataset are spaced out or clustered together. The spread is especially important when interpreting the variability within data, which is where measures like variance and standard deviation come in.
  • Low spread means the data points are close to the mean, resulting in a smaller variance and standard deviation.
  • High spread indicates that the data points are far from the mean, leading to larger variance and standard deviation values.
In the exercise at hand, selecting numbers like 0, 0, 10, and 10 results in a high spread with a standard deviation of 5. This is because the numbers chosen cover the whole range from 0 to 10, thus maximizing the data spread. Understanding how spread affects standard deviation helps clarify why data with varied numbers has a larger standard deviation.
Identical Numbers
Choosing identical numbers is a reliable way to achieve the smallest possible variance and standard deviation. This is because identical numbers have no variation from each other; thus, they are perfectly close to the mean.When you select numbers like 5, 5, 5, and 5, the calculations show:
  • The mean, \( \mu \), is 5 as all numbers are the same: \( \frac{5 + 5 + 5 + 5}{4} = 5 \).
  • Variance, \( \sigma^2 \), is zero because there is no deviation from the mean: \( \frac{(5-5)^2 + (5-5)^2 + (5-5)^2 + (5-5)^2}{4} = 0 \).
  • Thus, the standard deviation \( \sigma \) is \( \sqrt{0} = 0 \).
When choosing four identical numbers, you're effectively minimizing the spread of the dataset to nothing. This exercise demonstrates how such a choice always results in the smallest possible standard deviation, which is optimal when minimal data spread is desired.

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Most popular questions from this chapter

Returns on stocks. How well have stocks done over the past generation? The Wilshire 5000 index describes the average performance of all U.S. stocks. The average is weighted by the total market value of each company's stock, so think of the index as measuring the performance of the average investor. Here are the percent returns on the Wilshire 5000 index for the years from 19712015: 22 ? WILSHIRE $$ \begin{array}{lc|cc|cc} \hline \text { Year } & \text { Return } & \text { Year } & \text { Return } & \text { Year } & \text { Return } \\ \hline 1971 & 17.68 & 1986 & 16.09 & 2001 & -10.97 \\ \hline 1972 & 17.98 & 1987 & 2.27 & 2002 & -20.86 \\ \hline 1973 & -18.52 & 1988 & 17.94 & 2003 & 31.64 \\ \hline 1974 & -28.39 & 1989 & 29.17 & 2004 & 12.62 \\ \hline 1975 & 38.47 & 1990 & -6.18 & 2005 & 6.32 \\ \hline 1976 & 26.59 & 1991 & 34.20 & 2006 & 15.88 \\ \hline 1977 & -2.64 & 1992 & 8.97 & 2007 & 5.73 \\ \hline 1978 & 9.27 & 1993 & 11.28 & 2008 & -37.34 \\ \hline 1979 & 25.56 & 1994 & -0.06 & 2009 & 29.42 \\ \hline 1980 & 33.67 & 1995 & 36.45 & 2010 & 17.87 \\ \hline 1981 & -3.75 & 1996 & 21.21 & 2011 & 0.59 \\ \hline 1982 & 18.71 & 1997 & 31.29 & 2012 & 16.12 \\ \hline & & & & & \end{array} $$ $$ \begin{array}{lc|cc|ll} 1983 & 23.47 & 1998 & 23.43 & 2013 & 34.02 \\ \hline 1984 & 3.05 & 1999 & 23.56 & 2014 & 12.07 \\ \hline 1985 & 32.56 & 2000 & -10.89 & 2015 & -0.24 \\ \hline \end{array} $$ What can you say about the distribution of yearly returns on stocks?

The 2013-2014 roster of the Seattle Seahawks, winners of the 2014 NFL Super Bowl, included 10 defensive linemen and nine offensive linemen. The weights in pounds of the 10 defensive linemen were \(\begin{array}{llllllllll}311 & 254 & 297 & 260 & 323 & 242 & 300 & 252 & 303 & 274\end{array}\) The mean of these data is (a) \(281.60\). (b) \(282.50\). (c) \(285.50\).

Household assets. Once every three years, the Board of Governors of the Federal Reserve System collects data on household assets and liabilities through the Survey of Consumer Finances (SCF). \({ }^{15}\) Here are some results from the 2013 survey. (a) Transaction accounts, which include checking, savings, and money market accounts, are the most commonly held type of financial asset. The mean value of transaction accounts per household was \(\$ 270,100\), and the median value was \(\$ 94,500\). What explains the differences between the two measures of center? (b) The mean value of retirement accounts per household, which includes Individual Retirement Account (IRA) balances and certain employersponsored accounts, was \(\$ 99,040\) but the median value was \(\$ 0\). What does a median of \(\$ 0\) say about the percentage of households with retirement accounts?

If a distribution is skewed to the left, (a) the mean is less than the median. (b) the mean and median are equal. (c) the mean is greater than the median.

Behavior of the median. Place five observations on the line in the Mean and Median applet by clicking below it. (a) Add one additional observation without changing the median. Where is your new point? (b) Use the applet to convince yourself that when you add yet another observation (there are now seven in all), the median does not change, no matter where you put the seventh point. Explain why this must be true.
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