/*! 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 75 The Russian mathematician \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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The Russian mathematician \(\mathrm{P}\). L. Chebyshev (1821-1894) showed that for any data set and any constant \(k\) greater than \(1,\) at least \(1-\left(1 / k^{2}\right)\) of the data must lie within \(k\) standard deviations on either side of the mean \(A\). For example, when \(k=2\), this says that \(1-\frac{1}{4}=\frac{3}{4}\) (i.e., \(\left.75 \%\right)\) of the data must lie within two standard deviations of \(A\) (i.e., somewhere between \(A-2 \sigma\) and \(A+2 \sigma\) ). (a) Using Chebyshev's theorem, what percentage of a data set must lie within three standard deviations of the mean? (b) How many standard deviations on each side of the mean must we take to be assured of including \(99 \%\) of the data? (c) Suppose that the average of a data set is \(A\). Explain why there is no number \(k\) of standard deviations for which we can be certain that \(100 \%\) of the data lies within \(k\) standard deviations on either side of the \(\operatorname{mean} A\)

Short Answer

Expert verified
(a) Approximately 88.89% of the data must lie within three standard deviations (3σ) of the mean. (b) Approximately 10 standard deviations (10σ) on each side of the mean must be taken to guarantee including 99% of the data. (c) It's impossible to be sure that 100% of the data lies within any specific number of standard deviations from the mean, because outliers may exist outside this range.

Step by step solution

01

Solve part (a)

To solve part (a), substitute \(k = 3\) in Chebyshev's theorem formula \(1 - (1 / k^2)\), which gives the proportion of data lying within three standard deviations from the mean. That's, \(1 - (1/3^2) = 1 - (1/9) = 8/9 = 0.8888...\). As a percentage, this is approximately 88.89%.
02

Solve part (b)

To solve part (b), you need to arrange Chebyshev's theorem formula to solve for \(k\). Given that \(99\% = 0.99\) of the data lies within \(k\) standard deviations from the mean \(A\), you can set up the equation \(0.99 = 1 - (1 / k^2)\). Solve this equation for \(k\) to get \(k = \sqrt{1 / (1 - 0.99)} \approx 10\). Thus, you must take approximately 10 standard deviations on each side of the mean to be assured of including \(99\%\) of the data.
03

Solve part (c)

For part (c), recognize that Chebyshev's theorem does not guarantee that \(100\%\) of the data will be within any specific number of standard deviations from the mean \(A\). Because outliers may exist in a data set and fall outside of the approximate range given by the theorem, there is no number \(k\) such that \(100\%\) of the data lies within \(k\) standard deviations on either side of the mean \(A\).

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Key Concepts

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

Standard Deviation
Standard deviation is a crucial concept in statistics, providing insight into the variability or dispersion of a data set relative to its mean. It's represented by the symbol \( \sigma \). The standard deviation measures how spread out the numbers in your data set are from the mean, which is the average of all data points in the set.

To find the standard deviation, you follow these simple steps:
  • First, calculate the mean (average) of the data set.
  • Then, subtract this mean from each data point and square the result. This is called the squared deviation.
  • Next, find the mean of all these squared deviations.
  • Finally, take the square root of this mean. The result is the standard deviation.
This measurement is significant because it allows us to quantify the extent to which individual data points differ from the mean, making it easier to understand how typical or atypical specific results are in the context of the entire data set.
Additionally, standard deviation plays a key role in Chebyshev's theorem, which utilizes it to determine how much of the data lies within a certain range around the mean.
Data Distribution
Data distribution refers to how often each value in a data set occurs. In simple terms, it's a way of organizing data to show their frequency occurrences, assisting in understanding patterns or trends.

Chebyshev's theorem is particularly useful for understanding data distributions that may not follow a bell-shaped normal distribution. The theorem provides a way to make statements about the proportion of data within certain ranges, based on their standard deviations, even if the data isn't normally distributed.

In essence, Chebyshev's theorem states that for any number \( k \) greater than 1:
  • At least \(1 - \frac{1}{k^2}\) of the data lies within \(k\) standard deviations from the mean.
  • This is incredibly useful for all types of data distributions, as it sets a minimum percentage of data contained within these bounds, giving us a safety net in statistical analysis.
For example, if \(k = 3\), at least 88.89% of data points will be within three standard deviations of the mean, as shown in the step-by-step solution in the exercise.
Mathematical Proof
Mathematical proof is a logical argument demonstrating the truth of a proposition or theorem. In the realm of statistics, it is often used to establish the validity of certain theorems or formulas, such as Chebyshev's theorem.

Chebyshev's theorem doesn't rely on data being normally distributed, unlike many statistical tools. Instead, it provides guaranteed boundaries for data spread solely based on standard deviations from the mean. Proving Chebyshev's theorem involves leveraging inequalities and mathematical logic to establish that at least a certain amount of data is covered within defined standard deviation limits.

To understand why Chebyshev's theorem assures no specific \( k \) can encompass 100% of data, consider this:
  • Data can have outliers, extreme values outside most of the data's range. These outliers may lie far from the mean, skewing data spread.
  • The theorem's proof shows minimum data coverage but cannot account for every potential outlier, hence not assuring 100% containment.
Mathematically, because of these variations, there’s no \( k \) such that 100% of data lies within \( k \) standard deviations. Chebyshev's theorem recognizes the omnipresent potential for outliers, demonstrating why data can rarely be entirely predictable.

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Most popular questions from this chapter

Consider the data set \\{-5,7,4,8,2,8,-3,-6\\} (a) Find the five-number summary of the data set. (Hint: see Exercise 33 ). (b) Draw a box plot for the data set.

Refer to Table \(15-18,\) which shows the birth weights (in ounces) of the 625 babies born in Cleansburg hospitals in 2016. $$ \begin{array}{c|c|c|c|c|c} \begin{array}{c} \text { More } \\ \text { than } \end{array} & \begin{array}{c} \text { Less } \\ \text { than or } \\ \text { equal } \\ \text { to } \end{array} & \begin{array}{c} \text { Num- } \\ \text { ber of } \\ \text { babies } \end{array} & \begin{array}{c} \text { More } \\ \text { than } \end{array} & \begin{array}{c} \text { Less } \\ \text { than or } \\ \text { equal } \\ \text { to } \end{array} & \begin{array}{c} \text { Num- } \\ \text { ber of } \\ \text { babies } \end{array} \\ \hline 48 & 60 & 15 & 108 & 120 & 184 \\ \hline 60 & 72 & 24 & 120 & 132 & 142 \\ \hline 72 & 84 & 41 & 132 & 144 & 26 \\ \hline 84 & 96 & 67 & 144 & 156 & 5 \\ \hline 96 & 108 & 119 & 156 & 168 & 2 \end{array} $$ (a) Give the length of each class interval (in ounces). (b) Suppose that a baby weighs exactly 5 pounds 4 ounces. To what class interval does she belong? Describe the endpoint convention. (c) Draw the histogram describing the 2016 birth weights in Cleansburg using the class intervals given in Table \(15-18 .\)

Table \(15-21\) shows the relative frequencies of the scores of a group of students on a philosophy quiz. $$ \begin{array}{l|c|c|c|c|c} \text { Score } & 4 & 5 & 6 & 7 & 8 \\ \hline \begin{array}{l} \text { Relative } \\ \text { frequency } \end{array} & 7 \% & 11 \% & 19 \% & 24 \% & 39 \% \end{array} $$ (a) Find the average quiz score. (b) Find the median quiz score.

The purpose is to practice computing standard deviations the old fashioned way (by hand). Granted, computing standard deviations this way is not the way it is generally done in practice; a good calculator (or a computer package) will do it much faster and more accurately. The point is that computing a few standard deviations the old-fashioned way should help you understand the concept a little better. If you use a calculator or a computer to answer these exercises, you are defeating their purpose. Find the standard deviation of each of the following data sets. (a) \\{3,3,3,3\\} (b) \\{0,6,6,8\\} (c) \\{-6,0,0,18\\} (d) \\{6,7,8,9,10\\}

Table \(15-22\) shows the relative frequencies of the scores of a group of students on a 10 -point math quiz. $$ \begin{array}{l|c|c|c|c|c|c|c} \text { Score } & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \begin{array}{l} \text { Relative } \\ \text { frequency } \end{array} & 8 \% & 12 \% & 16 \% & 20 \% & 18 \% & 14 \% & 12 \% \end{array} $$ (a) Find the average quiz score rounded to two decimal places. (b) Find the median quiz score.

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