Do such questions appear regularly?This blew me out. Any simple way?
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vinay1983
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Do such questions appear regularly?This blew me out. Any simple way?
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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vipulgoyal
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As we can see the frequency distribution chart for X 10 appears one time, 20 appears 2 times and so on till 70 appear one time,
mean = [10+20+20+30+30+30+40+40+40+40+50+50+50+60+60+70]/16
we get 40, and here we can see in normal distribution median is also 40
so mean = median for X's distribution
using same calculation we"ll find mean = median for Z's distribution which is 40
Ans is E, we dont need to check for Y
In order to avoid calculation for Z, for quick visualize approach we can see graph is symmetrical around 40
mean = [10+20+20+30+30+30+40+40+40+40+50+50+50+60+60+70]/16
we get 40, and here we can see in normal distribution median is also 40
so mean = median for X's distribution
using same calculation we"ll find mean = median for Z's distribution which is 40
Ans is E, we dont need to check for Y
In order to avoid calculation for Z, for quick visualize approach we can see graph is symmetrical around 40
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Brent@GMATPrepNow
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Here's a slightly different approach.
Notice that distributions X and Z are each symmetrical.
For example, in both distributions, for every measurement of 10 there's a measurement of 70 (average =40).
For every measurement of 20 there's a measurement of 60 (average =40)
For every measurement of 30 there's a measurement of 50 (average =40)
So, in distributions X and Z, the mean (average) will be 40, and the median (middle-most value) will be 40
Answer: E
Cheers,
Brent
Notice that distributions X and Z are each symmetrical.
For example, in both distributions, for every measurement of 10 there's a measurement of 70 (average =40).
For every measurement of 20 there's a measurement of 60 (average =40)
For every measurement of 30 there's a measurement of 50 (average =40)
So, in distributions X and Z, the mean (average) will be 40, and the median (middle-most value) will be 40
Answer: E
Cheers,
Brent
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Brent@GMATPrepNow
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I wouldn't say that this is a frequently-occurring question type, but it is well within the scope of the GMAT.vinay1983 wrote: Do such questions appear regularly?
Cheers,
Brent
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Hi vinay1983,
The concepts in this question are likely to appear on the GMAT (whether in this format or some other).
You will see:
-Stats questions
-Questions about Mean and/or Median
-A Roman Numeral Question
The interesting thing about this question is that you don't have to calculate anything to solve it. If you understand the "patterns" behind groups of numbers, then you can see that 2 of the tables have a Mean that is equal to the Median. Based on the answer choices, you don't even have to consider the third table. THAT "pattern matching" ability is something that the GMAT will based several questions on, so it's a skill worth developing.
GMAT assassins aren't born, they're made,
Rich
The concepts in this question are likely to appear on the GMAT (whether in this format or some other).
You will see:
-Stats questions
-Questions about Mean and/or Median
-A Roman Numeral Question
The interesting thing about this question is that you don't have to calculate anything to solve it. If you understand the "patterns" behind groups of numbers, then you can see that 2 of the tables have a Mean that is equal to the Median. Based on the answer choices, you don't even have to consider the third table. THAT "pattern matching" ability is something that the GMAT will based several questions on, so it's a skill worth developing.
GMAT assassins aren't born, they're made,
Rich
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sana.noor
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looking at the graph pne can predict that its x and z because the values are equally distributed. so definitely mean and median of these two will be equal
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