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Re: The average (arithmetic mean) population in town X was recor [#permalink]
3
22,457. There is a simple shortcut for a change to an average. The figure for 2009 was recorded
as 22,478, but actually should have been recorded as 22,500, meaning 22 people in that year were not
counted. Thus, the sum should have been 22 higher when the average was originally calculated.
2000–2010, inclusive, is 11 years (subtract low from high and then add 1 to count an inclusive list of
consecutive numbers). When taking an average, divide the sum by the number of things being
averaged (in this case, 11). So the shortcut is to take the change to the sum and “spread it out” over all
of the values being averaged by dividing the change by the number of things being averaged.
Divide 22 by 11 to get 2. The average should have been 2 greater. Thus, the correct average for the
11-year period is 22,457.
Alternatively, use the traditional method: 22,455 × 11 years = 247,005, the sum of all 11 years’
recorded populations. Add the 22 uncounted people, making the corrected sum 247,027. Divide by 11
to get the corrected average: 22,457. (Note that while the traditional method is faster to explain, the
shortcut is faster to actually execute!)
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Re: The average (arithmetic mean) population in town X was recor [#permalink]
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This question is tricky.
First you need to work with the sum. it says there was an error in one of the year 2009 was 22478 and should have been 25000.
Second pull out the average formula
1- average before the edit was
22455 = (Sum/11)= and Sum= 11*22455= ‭247,005
We know the sum of the years is 247005 and should add change to it.
247005-22478+22500=‭247027‬ ==> this is the new sum after the edit.
finally the new average =247027/11=22457
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Re: The average (arithmetic mean) population in town X was recor [#permalink]
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