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USA Today (May 9,2006 ) published the accompanying weekday circulation numbers for the top 20 newspapers in the country for the 6 -month period ending March 31,2006: \(\begin{array}{rrrrr}2,272,815 & 2,049,786 & 1,142,464 & 851,832 & 724,242 \\\ 708,477 & 673,379 & 579,079 & 513,387 & 438,722 \\ 427,771 & 398,329 & 398,246 & 397,288 & 365,011 \\ 362,964 & 350,457 & 345,861 & 343,163 & 323,031\end{array}\) Explain why the average may not be the best measure of a typical value for this data set.

Short Answer

Expert verified
The average may not be the best measure of a typical value for this data set because it is influenced by the few high circulation numbers, causing it to be skewed higher. The median or mode may be better measures of central tendency in this scenario to give a more 'typical' value.

Step by step solution

01

Understanding the Mean

The mean, or average, is calculated by adding up all values in a data set and dividing by the number of values. This is commonly used as a measure of central tendency to find a typical value.
02

Identifying Outliers

Outliers are extreme values that deviate greatly from the other observations in a dataset. Looking at the newspaper circulation data, it can be seen that there are a few values which are significantly higher than the others, specifically: 2,272,815 and 2,049,786.
03

Impact of Outliers on Mean

Outliers can have a big impact on the mean, skewing it in their direction. This is because they contribute significantly more to the total sum when calculating the mean. In this case, the high circulation numbers of the top newspapers may increase the mean, making it higher than what we would intuitively see as 'typical'.
04

Considering Other Measures of Central Tendency

The median, which is the middle value when data is ordered from smallest to largest, is less affected by outliers and may be a better measure of a 'typical' value in this scenario. Similarly, the mode, which is the most frequently occurring value, could be considered too.

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Most popular questions from this chapter

Data on weekday exercise time for 20 females, consistent with summary quantities given in the paper "An Ecological Momentary Assessment of the Physical Activity and Sedentary Behaviour Patterns of University Students" (Health Education Journal [2010]: 116-125), are shown below. Female-Weekday \(\begin{array}{rrrrrr}10.0 & 90.6 & 48.5 & 50.4 & 57.4 & 99.6 \\\ 0.0 & 5.0 & 0.0 & 0.0 & 5.0 & 2.0 \\ 10.5 & 5.0 & 47.0 & 0.0 & 5.0 & 54.0 \\\ 0.0 & 48.6 & & & & \end{array}\) a. Calculate and interpret the values of the median and interquartile range. b. How do the values of the median and interquartile range for women compare to those for men calculated in the previous exercise?

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