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Find the average \(A\) and the median \(M\) of each data set. (a) \\{5,10,15,20,25,60\\} (b) \\{105,110,115,120,125,160\\}

Short Answer

Expert verified
For set (a), the average (A) is 22.5 and the median (M) is 17.5. For set (b), the average (A) is 122.5 and the median (M) is 117.5.

Step by step solution

01

Find the average of set (a)

To find the average (mean) of a data set, add all the numbers together and divide by the number of items in the set. For set (a), the sum of the numbers is 5 + 10 + 15 + 20 + 25 + 60, which is 135. There are 6 items in the set, so the average is 135 divided by 6.
02

Find the median of set (a)

To find the median, first arrange the data set in ascending order (if it isn't already), then locate the middle number. Since there are 6 numbers, the median will be the average of the third and fourth numbers, which are 15 and 20. The median is (15 + 20)/2.
03

Find the average of set (b)

Similar to step 1, for set (b), calculate the sum of the numbers, which is 105 + 110 + 115 + 120 + 125 + 160, equaling 735. There are 6 items in this set too, so the average is 735 divided by 6.
04

Find the median of set (b)

Just like in step 2, for set (b), there are 6 numbers, so find the average of the third and fourth items, which are 115 and 120. The median is (115 + 120)/2.

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Key Concepts

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

Mean Calculation
Calculating the mean, also known as the average, is a fundamental statistical measure used to find the central tendency of a data set. To compute the mean, you sum up all the numbers in the set and then divide by the count of these numbers.

For instance, consider a data set \(a\): \(\{5,10,15,20,25,60\}\). The sum of these numbers is 135. With six numbers in the set, the average is determined by dividing the sum by six, that is, \( \frac{135}{6} \), which gives us the mean value of 22.5.

This indicator is particularly useful in understanding the overall trend of a set of numbers, but it's important to note that the mean can be affected by outliers or values that are much different from the others in the set.
Median Calculation
While the mean calculates an average, the median provides the middle value of a sorted data series. It's less influenced by outliers than the mean and can thus provide a better sense of what is 'typical' for the data.

To calculate the median, you first arrange the data points in ascending order. If the number of values is odd, the median is the middle number. For an even number of values, such as in our data set \(a\), the median is the average of the two middle numbers. In this case, the third and fourth numbers are 15 and 20, so the median is \( \frac{15 + 20}{2} = 17.5 \). In the context of data set \(b\), which is also even-numbered, the median is calculated similarly, yielding a value of \( \frac{115 + 120}{2} = 117.5 \).
Data Set Analysis
Data set analysis involves various techniques to summarize and interpret collections of data. Cronological organization, summarization, and the application of statistical measures like mean and median are part of this process.

Averages such as mean and median are essential tools in data set analysis. They give a quick snapshot of the data and can indicate trends or anomalies. For example, when comparing sets \(a\) and \(b\) from the exercise, the presence of the number 60 in set \(a\) and 160 in set \(b\) could significantly influence the mean and point towards the existence of outliers.

Understanding such nuances is crucial for accurate analysis. Visualization techniques such as box plots can also help to illustrate the central tendency and dispersion of data, thus aiding in effective data set analysis.
Mathematical Averages
Mathematical averages are a cornerstone of data interpretation, providing a simplified representation of complex data sets. The two most common types of averages in mathematics are math mean and median, each highlighting different aspects of the data.

The mean gives a mathematical balance point of the data, but can be skewed by extreme values. On the other hand, the median offers a midpoint that isn't swayed by outliers. It's essential to select the appropriate average to analyze a given set of data correctly, and sometimes it's beneficial to use both to gain a comprehensive understanding of the data's characteristics.

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Most popular questions from this chapter

For each data set, find the 10 th and the 25 th percentiles. (a) \(\\{1,2,3, \ldots, 49,50,50,49, \ldots, 3,2,1\\}\) (b) \(\\{1,2,3, \ldots, 49,50,49, \ldots, 3,2,1\\}\) (c) \(\\{1,2,3, \ldots, 49,49, \ldots, 3,2,1\\}\)

Exercises 73 and 74 refer to histograms with unequal class intervals. When sketching such histograms, the columns must be drawn so that the frequencies or percentages are proportional to the area of the column. Figure 21 illustrates this. If the column. over class interval 1 represents \(10 \%\) of the population, then the column over class interval \(2,\) also representing \(10 \%\) of the population, must be one-third as high, because the class interval is three times as large (Fig. 21 ).. If the height of the column over the class interval \(20-30\) is one unit and the column represents \(25 \%\) of the population, then (a) how high should the column over the interval \(30-35\) be if \(50 \%\) of the population falls into this class interval? (b) how high should the column over the interval \(35-45\) be if \(10 \%\) of the population falls into this class interval? (c) how high should the column over the interval \(45-60\) be if \(15 \%\) of the population falls into this class interval?

Table 16 shows the percentage of U.S. working married couples in which the wife's income is higher than the husband's \((1999-2009)\) (a) Draw a pictogram for the data in Table 16 . Assume you are trying to convince your audience that things are looking great for women in the workplace and that women's salaries are catching up to men's very quickly. (b) Draw a different pictogram for the data in Table 16 , where you are trying to convince your audience that women's salaries are catching up with men's very slowly. $$ \begin{array}{|l|c|c|c|c|c|c|}= \hline \text { Year } & 1999 & 2000 & 2001 & 2002 & 2003 & 2004 \\ \hline \text { Percent } & 28.9 & 29.9 & 30.7 & 31.9 & 32.4 & 32.6 \\ \hline \text { Year } & 2005 & 2006 & 2007 & 2008 & 2009 & \\\ \hline \text { Percent } & 33.0 & 33.4 & 33.5 & 34.5 & 37.7 & \end{array} $$

(a) Show that if \(\left\\{x_{1}, x_{2}, x_{14} \ldots x_{N}\right\\}\) is a data set with mean \(A\) and standard deviation \(\boldsymbol{\sigma},\) then \(\sigma \sqrt{N} \geq\left|x_{i}-A\right|\) for every data value \(x_{i}\) (b) Use (a) to show that every data value is bigger than or equal to \(A-\sigma \sqrt{N}\) and smaller than or equal to \(A+\sigma \sqrt{N}\) (i.e., for every data value \(x\), \(A-\sigma \sqrt{N} \leq x \leq A+\sigma \sqrt{N}\).

Table 17 shows the percentage of U.S. workers who are members of unions \((2000-2011)\). (a) Draw a pictogram for the data in Table \(17 .\) Assume you are trying to convince your audience that unions are holding their own and that the percentage of union members in the workforce is steady. (b) Draw a different pictogram for the data in Table 17 . where you are trying to convince your audience that there is a steep decline in union membership in the U.S. workforce. $$ \begin{array}{|l|c|c|c|c|c|c|} \hline \text { Year } & 2000 & 2001 & 2002 & 2003 & 2004 & 2005 \\ \hline \text { Percent } & 13.4 & 13.3 & 13.3 & 12.9 & 12.5 & 12.5 \\ \hline \text { Year } & 2006 & 2007 & 2008 & 2009 & 2010 & 2011 \\ \hline \text { Percent } & 12.0 & 12.1 & 12.4 & 12.3 & 11.9 & 11.8 \end{array} $$

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