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List X and list Y each contain 60 numbers. Frequency distrib

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List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 08 Dec 2015, 08:16
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List X and list Y each contain 60 numbers. Frequency distributions for each list are given above. The average (arithmetic mean) of the numbers in list X is
2.7, and the average of the numbers in list Y is 7.1. List Z contains 120 numbers: the 60 numbers in list X and the 60 numbers in list Y.

Quantity A
Quantity B
The average of the 120 numbers in list Z
The median of the 120 numbers in list Z


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.




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Question: 2
Page: 150
Difficulty: medium
[Reveal] Spoiler: OA

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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 08 Dec 2015, 08:34
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Solution

This kind of question should be attacked with an upfront work. Students can be misleading by frequency or else. But dive into the problem and see the real gist of it

List Z is the sum of list X and Y. So the average of Z in the average of the single averages of X and Y : \(\frac{2.7+7.1}{2}=4.9\)

To compare the Average and the median, we have to calculate the median of Z. Z is composed of 120 numbers, so the median is the average of the two middle numbers.
X have numbers from 1 to 5, and Y from 6 to 9. As such, the two middle numbers of Z are 5 and 6. the median is \(\frac{6+5}{2}=5.5\)

So the correct answer is \(B\)
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 10 Dec 2016, 09:06
why median is 5+6 .....not only 5 ??
they are not even numbers ?
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 10 Dec 2016, 09:52
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awamy wrote:
why median is 5+6 .....not only 5 ??
they are not even numbers ?


Median is the exact mid point when a list of numbers are arranged in increasing order.

So in 1, 2, 3, 5, 7 median is 3.

But in 1, 2, 3, 5 median is \(\frac{2+3}{2}=2.5\)
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 10 Dec 2016, 09:55
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ohhh, yesss i didn't notice that there is no value (4)
thnx a lot for ur prompt reply
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 20 Dec 2017, 20:01
Carcass wrote:
Solution

This kind of question should be attacked with an upfront work. Students can be misleading by frequency or else. But dive into the problem and see the real gist of it

List Z is the sum of list X and Y. So the average of Z in the average of the single averages of X and Y : \(\frac{2.7+7.1}{2}=4.9\)

To compare the Average and the median, we have to calculate the median of Z. Z is composed of 120 numbers, so the median is the average of the two middle numbers.
X have numbers from 1 to 5, and Y from 6 to 9. As such, the two middle numbers of Z are 5 and 6. the median is \(\frac{6+5}{2}=5.5\)

So the correct answer is \(B\)


Hi Carcass,
Shouldn't the median be (6+6)/2= 6?
the median should be measure (order60+order61)/2
I apologized because I don't know how to explain better
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 21 Dec 2017, 13:45
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wongpcla wrote:
Carcass wrote:

Hi Carcass,
Shouldn't the median be (6+6)/2= 6?
the median should be measure (order60+order61)/2
I apologized because I don't know how to explain better


Hi

The 60th term (order 60) has value 5 and and the 61st term (order 61) has value 6 so the median is average of both 5 and 6. Hence 5.5.
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 02 Apr 2018, 12:09
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The answer should be B, I believe its keyed in incorrectly.
Thank you again for putting this together!!
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 01 Oct 2018, 11:45
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awamy wrote:
why median is 5+6 .....not only 5 ??
they are not even numbers ?

Median for odd number = (n+1/2)th
Median for even number = 1/2{(n/2)th+(n/2+1)th}
so, 5 and 6 are median
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 02 Oct 2019, 10:58
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That is a clever solution, Carcass. I would just add for anyone who missed it, we can use the "average of the single averages" because each set has an equal number of values; in other words, their weight is equal. If the sets had unequal numbers of values, for example if the first set had 59 values and the second had 60, then we couldn't average the averages.

Having said this, when we come across a problem like this, where we can use your method it will save a lot of time, versus calculating a weighted mean, or just calculating a total sum for both sets using n and the mean (ex. 60x2.7 + 60x7.1) and then dividing to find the aggregate mean ( [60x2.7 + 60x7.1]/120 = 4.9).
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink] New post 02 Oct 2019, 11:31
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Re: List X and list Y each contain 60 numbers. Frequency distrib   [#permalink] 02 Oct 2019, 11:31
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