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04 Nov 2019, 03:07
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If it were discovered that the value of imports shown for 2007 was incorrect and should have been $5 billion instead, then the average (arithmetic mean) value of imports per year for the 10 years shown would have been approximately how much less?
A $200 million
B $50 million
C $20 million
D $7 million
E $5 million
06 Nov 2019, 14:54
Kudos
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To find the decrease in average, we must calculate the average with the incorrect value for 2007 and the correct value for 2007 (in billions)
\(\frac{(3+3+3+4+4+5+6+7+5+9)}{10}\)=$4.9 billion
With changed data point for 2007:
\(\frac{(3+3+3+4+4+5+6+5+5+9)}{10}\)=$4.7 billion
The difference is .2 billion, or $200 million
Ans=A
\(\frac{(3+3+3+4+4+5+6+7+5+9)}{10}\)=$4.9 billion
With changed data point for 2007:
\(\frac{(3+3+3+4+4+5+6+5+5+9)}{10}\)=$4.7 billion
The difference is .2 billion, or $200 million
Ans=A
07 Nov 2019, 04:08
Kudos
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The value of imports shown for 2007 is incorrect and should have been $5 billion instead. We can see in the graph that it is incorrectly valued at 7 billion.
So it is, after correction, for the year 2007, 7 billion - 5 billion = 2 billion less.
In the operation of arithmetic mean, we add up differents values divided by the same denominator, which is the number of elements taken into consideration in the arithmetic mean.
So, for the arithmetic mean of 10 years, we add different years imports' value in dollars, all divided by 10.
We want to remove 2 billion from the value of 2007 imports.
So, New Arithmetic mean = (Year 2000 + Year 2001 + Year 2002 + Year 2003 + Year 2004 + Year 2005 + Year 2006 + Year 2007 - 2 billion + Year 2008 + Year 2009)/10
= (Year 2000 + Year 2001 + Year 2002 + Year 2003 + Year 2004 + Year 2005 + Year 2006 + Year 2007 + Year 2008 + Year 2009)/10 - 2 billion/(10)
= (Year 2000 + Year 2001 + Year 2002 + Year 2003 + Year 2004 + Year 2005 + Year 2006 + Year 2007 + Year 2008 + Year 2009)/10 - 200 million
= Old Arithmetic mean - 200 million
We conclude that the average (arithmetic mean) value of imports per year for the 10 years shown would have been approximately 2 billion divided by 10 equals 200 million less.
So it is, after correction, for the year 2007, 7 billion - 5 billion = 2 billion less.
In the operation of arithmetic mean, we add up differents values divided by the same denominator, which is the number of elements taken into consideration in the arithmetic mean.
So, for the arithmetic mean of 10 years, we add different years imports' value in dollars, all divided by 10.
We want to remove 2 billion from the value of 2007 imports.
So, New Arithmetic mean = (Year 2000 + Year 2001 + Year 2002 + Year 2003 + Year 2004 + Year 2005 + Year 2006 + Year 2007 - 2 billion + Year 2008 + Year 2009)/10
= (Year 2000 + Year 2001 + Year 2002 + Year 2003 + Year 2004 + Year 2005 + Year 2006 + Year 2007 + Year 2008 + Year 2009)/10 - 2 billion/(10)
= (Year 2000 + Year 2001 + Year 2002 + Year 2003 + Year 2004 + Year 2005 + Year 2006 + Year 2007 + Year 2008 + Year 2009)/10 - 200 million
= Old Arithmetic mean - 200 million
We conclude that the average (arithmetic mean) value of imports per year for the 10 years shown would have been approximately 2 billion divided by 10 equals 200 million less.
