Mike, Ana, Tiffany, Josh, And Annie Are Heading To The Store
Mike Ana Tiffany Josh And Annie Are Heading To The Store To Get Som
Mike, Ana, Tiffany, Josh, and Annie are heading to the store to get some snacks. Initially, Mike has $1, Ana has $2, Tiffany has $3, Josh has $4, and Annie has $5. The task is to find the average (mean) amount of cash the five kids have and the median, showing all work and calculations.
A few days later, Annie's family wins the lottery, and the kids go to the store again. This time, Mike has $1, Ana has $2, Tiffany has $3, Josh has $4, and Annie has a total of $5,000. The same calculations are to be performed: find the mean and median, and discuss how these measures have changed from the previous situation.
Finally, an analysis is required to determine which of the two—mean or median—is a better reflection of how much money they have as a distribution and why.
Paper For Above instruction
In this analysis, we examine how the average and median amounts of money held by the children change under two different conditions and discuss which measure more accurately reflects their monetary situation as a distribution. The initial data set suggests a relatively evenly spread amount of cash, while the latter reflects a considerable disparity introduced by Annie's lottery win.
Part 1: Initial Scenario Calculation
Initially, the five children—Mike, Ana, Tiffany, Josh, and Annie—hold $1, $2, $3, $4, and $5 respectively. To find the mean, we sum all amounts and divide by five:
\[
\text{Mean} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3
\]
The median, the middle value when the amounts are ordered, is the third value in the sorted list: $1, $2, $3, $4, $5. The median is $3.
This initial scenario showcases a balanced distribution with the mean and median coinciding at $3.
Part 2: Post-Lottery Scenario Calculation
In the second scenario, Annie's cash dramatically increases to $5,000, while the other children’s amounts remain unchanged: Mike has $1, Ana $2, Tiffany $3, Josh $4, and Annie $5,000.
The sum of their cash holdings is:
\[
1 + 2 + 3 + 4 + 5000 = 5010
\]
The mean is:
\[
\text{Mean} = \frac{5010}{5} = 1002
\]
The sorted list is $1, $2, $3, $4, $5,000. The median, being the middle value, is the third in the list, which is now $3.
This dramatic increase in the mean, from 3 to 1002, demonstrates the effect of the large lottery win, while the median remains at $3.
Comparison and Analysis of Mean and Median Changes
The mean increased significantly in the second scenario, reflecting sensitivity to the extremely high value of Annie’s lottery cash. Conversely, the median remained unchanged because it depends solely on the middle value, unaffected by outliers or extreme values.
This discrepancy indicates that the mean is heavily influenced by outliers—large or small values that deviate substantially from the rest of the data—whereas the median better represents the typical amount of cash held by most children.
Which Measure Better Reflects the Distribution?
Given the substantial disparity introduced by Annie’s lottery winnings, the median provides a more accurate picture of the typical amount of money held by the children in this scenario. The median isn't skewed by Annie’s sudden increase, which pulls the mean upward dramatically. In real-world contexts where outliers are present, the median offers a more robust central tendency measurement, reflecting the common experience more reliably than the mean.
Conclusion
The analysis highlights how outliers can distort the mean, making it less representative of the typical data point in skewed distributions. Conversely, the median offers stability amid such outliers, making it a more meaningful statistic for understanding central tendency when data include extreme values. The huge lottery win exemplifies these differences clearly, emphasizing the importance of choosing the appropriate measure based on the distribution's nature.
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