sandy wrote:
Attachment:
The attachment #GREpracticequestion The graph above shows the frequency distribution of 50.jpg is no longer available
The graph above shows the frequency distribution of 50 integer values varying from 1 to 6.
Quantity A |
Quantity B |
The average (arithmetic mean) of the 50 values |
The median of the 50 values |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
From the diagram, we can see that the 50 values consist of:
Two 1's, four 2's, seven 3's, ten 4's, sixteen 5s and eleven 6's.
So, the MEAN \(\frac{= (2)(1) + (4)(2) + (7)(4) + (16)(5) + (11)(6)}{50}\)
\(= \frac{2 + 8 + 28 + 80 + 66}{50}\)
\(=\frac{184}{50}\)
\(3.68\)
-----------------------------------
Since we have an EVEN number of values (50 in total) the MEDIAN will be the average of the 2 middlemost values.
So, when all 50 values are arranged in
ascending order, the two middlemost values will be the
25th and
26th values.
Let's list the values in ascending order:
{1,1, 2,2,2,2, 3,3,3,3,3,3,3, 4,4,4,4,4,4,4,4,4,4 ...(we've listed 23 values so far. So, keep going...)
{1,1, 2,2,2,2, 3,3,3,3,3,3,3, 4,4,4,4,4,4,4,4,4,4, 5,
5,
5,5,5,5,5,5,5,5,...}
We can stop here.
The two middlemost values are
5 and
5So, the MEDIAN \(= \frac{5+5}{2} = 5\)
-----------------------------------
We get:
Quantity A: \(3.68\)
Quantity B: \(5\)
Answer: B
Cheers,
Brent
Attachment:
#GREpracticequestion The graph above shows the frequency distribution of 50.jpg [ 11.64 KiB | Viewed 1857 times ]
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Brent Hanneson – Creator of greenlighttestprep.com
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