Question 314246: The mean household incomes of randomly selected residents of Temecula City, CA in U.S. dollars are listed.
$62,050 $59,250 $65,100 $66,000 $32,000
$69,150 $32,000 $61,120 $61,000
From this data set, find the sample mean, median, mode and range for the household incomes. Round each to the nearest dollar if necessary. Include units with each answer.
Which of the numbers you calculated is a better measure of the central tendency of the incomes given? Give a specific statistical reason for your conclusion.
Answer by Jstrasner(112)   (Show Source):
You can put this solution on YOUR website! First, the mean is the average of all of the incomes combined together.
This means all of the numbers (in this case incomes) added together and divided by the number of incomes (in this case, 9)
Therefore:
((32,000)+(32,000)+(59,250)+(61,000)+(61,120)+(62,050)+(65,100)+(66,000)+(69,150))/9= 56,407.777777 which you round up to 56,408, so the mean of the incomes is $56,408.
The median of the incomes is the middle (literally) number, or income. You need to put all of the numbers in order from smallest to biggest. Then you cross off numbers on both ends of your list until you get the middle number. Therefore:
Out of $32,000 $32,000 $59,250 $61,000 $61,120 $62,050 $65,100 $66,000 $69,150 the median income is $61,120.
The mode of the incomes is the number that appears the most amount of times. In this case $32,000 is the mode because it appears twice in the list of incomes.
The range of numbers is the difference between the greatest and the smallest numbers. Therefore:
$69,150-$32,000= $37,150
The last part of the question asks for the best answer that measures the central tendency of the incomes. The answer would be the median because the mode and the range answers are too low to properly represent the income tendency of the incomes given. The mean is no the best answer either because the majority of the incomes are in the 60,000s.
Therefore, the answer would be the median, which is $61,120.
I hope this helps.
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