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Bunuel
In a set of twenty numbers, 19 of the 20 numbers are between 40 and 50. Is the median greater than the mean?

(1) The standard deviation is greater than 15

(2) The 20th number is greater than 100


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MAGOOSH OFFICIAL SOLUTION:

For a symmetrical distribution, the mean = median, and if the median is close to symmetry, the mean and the median are close in value. When the distribution of numbers is radical asymmetrical, with one outlier or several outliers on only one side of the distribution, then the mean is pulled in the direction of the outliers. The median, resistant to outliers, stays in the middle of the majority of numbers, but the mean is sensitive to outliers, gets pulled in their direction. High outliers pull the mean up, and low outliers pull the mean down.

Statement #1: the standard deviation is greater than 15

This tells us that there’s large variation, suggesting that the 20th number is far away from the other 19, but far away in which direction? Much higher or much lower than the rest of the numbers? We don’t know. A high outlier would pull the mean up, and a low outlier would pull the mean down. Here, we know we have an outlier, but we don’t know its directions, so we don’t know in which direction the mean is affected. We cannot answer the question. This statement, alone and by itself, is not sufficient.

Statement #2: the 20th number is greater than 100

Now, we know that the outlier is a high outlier, much bigger than the other numbers in the set. A high outlier pulls the mean up, away from the median, so the mean is higher than the median. We can give a definitive “yes” answer to the prompt question. This statement, alone and by itself, is sufficient.

Answer = (B)
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If we consider this following scenario
First 9 numbers: 41 X 9 = 369 [ sum]
The second set of 10 numbers : 49 X 10 = 490[sum]

case- I
the last number is assumed to be 101
so Mean = 48
Median = 49
Median > mean.
---
Case - II:
The last number is assumed to be 10,000
now, we can safely conclude that mean>median.
--
So, How B is the correct answer ? Where I am going wrong..

Edit 1 - grammatical error/spelling error
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godot53
If we consider this following scenario
First 9 numbers: 41 X 9 = 369 [ sum]
The second set of 10 numbers : 49 X 10 = 490[sum]

case- I
the last number assumed to be 101
so Mean = 48
Median = 49
Median > mean.
---
Case - II:
The last number is assumed is 10,000
now, we can safely conclude that mean>median.
--
So, How is B the correct answer ? Where I am going wrong..

Hi godot53,

I think your example is perfectly fine. I don't see any error in it. Let others comment on it.

Thank you.
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Statement 1: No information is given on the 20th value making it impossible to determine the limit of the set, and thus the mean cannot be determined.

Statement 2: We are given that the limit of the set is greater than 100. This is sufficient as we can answer YES/NO on whether the median is greater than the mean.
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Hello Bunuel : It would be great if you can consider this.
godot53
If we consider this following scenario
First 9 numbers: 41 X 9 = 369 [ sum]
The second set of 10 numbers : 49 X 10 = 490[sum]

case- I
the last number is assumed to be 101
so Mean = 48
Median = 49
Median > mean.
---
Case - II:
The last number is assumed to be 10,000
now, we can safely conclude that mean>median.
--
So, How B is the correct answer ? Where I am going wrong..

Edit 1 - grammatical error/spelling error
I understand that there is some problem with this question.

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