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Median of a set containing unknown values? [#permalink]
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18 Jan 2014, 12:28
Hi,
I was reading mgmat quant statistics chapter wherein i am not able to understand the median of an unknown value type. can somebody please tell how to go about in these set of questions?
anuj



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Re: Median of a set containing unknown values? [#permalink]
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18 Jan 2014, 22:24
anu1706 wrote: Hi,
I was reading mgmat quant statistics chapter wherein i am not able to understand the median of an unknown value type. can somebody please tell how to go about in these set of questions?
anuj Anuj: More than happy to help out  the thing that is unique about medians is that you often don't need to know all of the numbers in a set to still figure out what the median is. For example, if I gave you this set of numbers (x, 4, 5, 7, 7, 9, 11) and you are told that they are all integers  no matter what 'x' is  the median is still 7. If x is 0, median is 7. If x is 100, median is 7. if x is 7.. guess what.. median is 7!! Does that answer your question? Brian
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Re: Median of a set containing unknown values? [#permalink]
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19 Jan 2014, 02:27
brianlange77 wrote: anu1706 wrote: Hi,
I was reading mgmat quant statistics chapter wherein i am not able to understand the median of an unknown value type. can somebody please tell how to go about in these set of questions?
anuj Anuj: More than happy to help out  the thing that is unique about medians is that you often don't need to know all of the numbers in a set to still figure out what the median is. For example, if I gave you this set of numbers (x, 4, 5, 7, 7, 9, 11) and you are told that they are all integers  no matter what 'x' is  the median is still 7. If x is 0, median is 7. If x is 100, median is 7. if x is 7.. guess what.. median is 7!! Does that answer your question? Brian Sorry but that not precisely my question was!! i can understand median is middle number in odd and avg. of 2 middle in even set of data points. What I want to ask is how unknown value will effect the median of a set..Please explain what this means.: Extract from manhattan "Consider the unordered set {x, 2, 5, 11, 11, 12, 33}. No matter whether x is less than 11, equal to 11, or greater than 11, the median of the resulting set will be 11. Why?By contrast, the median of the unordered set {x, 2, 5, 11, 12, 12, 33} depends on x. If x is 11 or less, the median is 11. If x is between 11 and 12, the median is x. Finally, if x is 12 or more, the median is 12. Explain!..



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Re: Median of a set containing unknown values? [#permalink]
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20 Jan 2014, 00:10
anu1706 wrote: Extract from manhattan "Consider the unordered set {x, 2, 5, 11, 11, 12, 33}. No matter whether x is less than 11, equal to 11, or greater than 11, the median of the resulting set will be 11. Why? By contrast, the median of the unordered set {x, 2, 5, 11, 12, 12, 33} depends on x. If x is 11 or less, the median is 11. If x is between 11 and 12, the median is x. Finally, if x is 12 or more, the median is 12. Explain!.. Notice how the two given sets are different. Ignoring x from the sets, the first set has two 11s in the middle (3rd and 4th terms) while the second one has an 11 and a 12 in the middle. Consider {2, 5, 11, 11, 12, 33} We need to add a term 'x' in this set and then find out the median. x can take any value and to find the median, we will put x in the set such that all terms are in ascending order. The middle term (fourth term) of this set will be the median. Depending on the value of x, several scenarios are possible. Let's give some values to x and see the various scenarios. x = 1; { 1, 2, 5, 11, 11, 12, 33} (median 11) x = 3; {2, 3, 5, 11, 11, 12, 33} (median 11) x = 9; {2, 5, 9, 11, 11, 12, 33} (median 11) x = 11; {2, 5, 11, 11, 11, 12, 33} (the only way x can be in the middle of the two 11s is if it is 11 too. x = 12; {2, 5, 11, 11, 12, 12, 33} (median 11) x = 25; {2, 5, 11, 11, 12, 25, 33} (median 11) x = 50; {2, 5, 11, 11, 12, 33, 50} (median 11) Fourth term will be the median. The only time the fourth term (middle term) will be x will be when x is 11. Hence, whatever the case, the median will always be 11. Consider {2, 5, 11, 12, 12, 33} Again, depending on the value of x, several scenarios are possible. Let's give some values to x and see the various scenarios. { 1, 2, 5, 11, 12, 12, 33} (median 11) {2, 3, 5, 11, 12, 12, 33} (median 11) {2, 5, 9, 11, 12, 12, 33} (median 11) {2, 5, 11, 11.2, 12, 12, 33} (if x is in the middle of 11 and 12, it can take many values such as 11.2/11.7 etc and that will be the median) {2, 5, 11, 12, 12, 12, 33} (median 12) {2, 5, 11, 12, 12, 25, 33} (median 12) {2, 5, 11, 12, 12, 33, 50} (median 12)
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Re: Median of a set containing unknown values? [#permalink]
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20 Jan 2014, 00:50
VeritasPrepKarishma wrote: anu1706 wrote: Extract from manhattan "Consider the unordered set {x, 2, 5, 11, 11, 12, 33}. No matter whether x is less than 11, equal to 11, or greater than 11, the median of the resulting set will be 11. Why? By contrast, the median of the unordered set {x, 2, 5, 11, 12, 12, 33} depends on x. If x is 11 or less, the median is 11. If x is between 11 and 12, the median is x. Finally, if x is 12 or more, the median is 12. Explain!.. Notice how the two given sets are different. Ignoring x from the sets, the first set has two 11s in the middle (3rd and 4th terms) while the second one has an 11 and a 12 in the middle. Consider {2, 5, 11, 11, 12, 33} We need to add a term 'x' in this set and then find out the median. x can take any value and to find the median, we will put x in the set such that all terms are in ascending order. The middle term (fourth term) of this set will be the median. Depending on the value of x, several scenarios are possible. Let's give some values to x and see the various scenarios. x = 1; { 1, 2, 5, 11, 11, 12, 33} (median 11) x = 3; {2, 3, 5, 11, 11, 12, 33} (median 11) x = 9; {2, 5, 9, 11, 11, 12, 33} (median 11) x = 11; {2, 5, 11, 11, 11, 12, 33} (the only way x can be in the middle of the two 11s is if it is 11 too. x = 12; {2, 5, 11, 11, 12, 12, 33} (median 11) x = 25; {2, 5, 11, 11, 12, 25, 33} (median 11) x = 50; {2, 5, 11, 11, 12, 33, 50} (median 11) Fourth term will be the median. The only time the fourth term (middle term) will be x will be when x is 11. Hence, whatever the case, the median will always be 11. Consider {2, 5, 11, 12, 12, 33} Again, depending on the value of x, several scenarios are possible. Let's give some values to x and see the various scenarios. { 1, 2, 5, 11, 12, 12, 33} (median 11) {2, 3, 5, 11, 12, 12, 33} (median 11) {2, 5, 9, 11, 12, 12, 33} (median 11) {2, 5, 11, 11.2, 12, 12, 33} (if x is in the middle of 11 and 12, it can take many values such as 11.2/11.7 etc and that will be the median) {2, 5, 11, 12, 12, 12, 33} (median 12) {2, 5, 11, 12, 12, 25, 33} (median 12) {2, 5, 11, 12, 12, 33, 50} (median 12) Thanks for the insight. can you please help in cementing the concept with the help of a official question please.




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