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List X and list Y each contain 60 numbers. Frequency distrib [#permalink]
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. Practice Questions Question: 2 Page: 150 Difficulty: medium
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink]
08 Dec 2015, 08:34
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SolutionThis 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]
10 Dec 2016, 09:06
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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]
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]
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]
20 Dec 2017, 20:01
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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]
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]
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]
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]
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). This is also a good time to remind people that the median is the (n+1)/2 th value . what I mean by this is: . when whole number result: the median is that number in the list . when decimal number result: the median is the average of the number at that place and the next place in this case (120+1)/2 > 60.5th value > means the median is the average of the 60th and 61st value, in this case 5 and 6, so 5+6/2 = 5.5 I think this is more obvious if you think of the frequency table as shown in the attached graphic. You can see there are 10 values that are 1, up to the 30th value is 2, up to the 48th value is 3, up to the 60th value is 5, up to the 84th value is 24... So, you can see that 60th value is 5 and the 61st value is 6, so again, the median is 5+6/2 = 5.5 My explanation may not be completely clear at first glance, but I think if you think about it, you'll see what I mean.
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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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gat aach



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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink]
27 Apr 2020, 13:12
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sandy wrote: 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 \(\frac{2+3}{2}=2.5\) In that case List Y has median (7+8)/ 2, no?
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink]
27 Apr 2020, 13:43
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Farina wrote: sandy wrote: 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 \(\frac{2+3}{2}=2.5\) In that case List Y has median (7+8)/ 2, no? If we take List Z and arrange all 120 numbers in ascending order, then 5 and 6 are the two middlemost terms. That is, the 60th value is 5 and the 61st value is 6. So, the median = (5+6)/2 = 5.5 Does that help? Cheers, Brent ASIDE: List Z: 1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,5,5,5,5,5,5,5,5,5,5,5, 5, 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9}
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink]
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GreenlightTestPrep wrote: Farina wrote: sandy wrote: 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 \(\frac{2+3}{2}=2.5\) In that case List Y has median (7+8)/ 2, no? If we take List Z and arrange all 120 numbers in ascending order, then 5 and 6 are the two middlemost terms. That is, the 60th value is 5 and the 61st value is 6. So, the median = (5+6)/2 = 5.5 Does that help? Cheers, Brent ASIDE: List Z: 1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,5,5,5,5,5,5,5,5,5,5,5, 5, 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9} Yes thank you. I was considering only list y, whereas we have to consider list z
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Re: List X and list Y each contain 60 numbers. Frequency distrib [#permalink]
28 Apr 2020, 11:48
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Farina wrote: Yes thank you. I was considering only list y, whereas we have to consider list z Funny you should say that. I was doing the exact same thing when I started. Then I had to go back and start over.
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