Tricky stats question

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Tricky stats question

by Mo2men » Wed Nov 23, 2016 2:32 am
The range of the heights of the female students in a certain class is 30 centimeters, and the range of the heights of the male students in the class is 40 centimeters. What is the range of the heights of all the students in the class?

(1) The average (arithmetic mean) height of the male students in the class is 10 centimeters greater than the average height of the female students in the class.

(2) The shortest female student in the class is 15 centimeters shorter than the shortest male student in the class.

OA:B

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by GMATGuruNY » Wed Nov 23, 2016 6:58 am
Mo2men wrote:The range of the heights of the female students in a certain class is 30 centimeters, and the range of the heights of the male students in the class is 40 centimeters. What is the range of the heights of all the students in the class?

(1) The average (arithmetic mean) height of the male students in the class is 10 centimeters greater than the average height of the female students in the class.

(2) The shortest female student in the class is 15 centimeters shorter than the shortest male student in the class.
Statement 1:
Let the average for 3 male students = 30, implying that the sum of the male heights = 3*30 = 90.
Let the average for 3 female students = 20, implying that the sum of the female heights = 3*20 = 60.

Case 1: shortest male = 10, shortest female = 10
Males:
Since range = 40, tallest male = 10+40 = 50.
Median male = sum - shortest - tallest = 90-10-50 = 30.
Females:
Since range = 30, tallest female = 10+30 = 40.
Median female = sum - shortest - tallest = 60-10-40 = 10.
In this case, the range for the whole class = (tallest male) - (shortest female) = 50-10 = 40.

Case 2: shortest male = 15, shortest female = 10
Males:
Since range = 40, tallest male = 15+40 = 55.
Median male = sum - shortest - tallest = 90-15-55 = 20.
Females:
Since range = 30, tallest female = 10+30 = 40.
Median female = sum - shortest - tallest = 60-10-40 = 10.
In this case, the range for the whole class = (tallest male) - (shortest female) = 55-10 = 45.

Since the range for the whole class can be different values, INSUFFICIENT.

Statement 2:
Case 1: shortest female = 1 centimeter, shortest male = 16 centimeters
Since the range of the male students = 40, tallest male = 16+40 = 56 centimeters.
Resulting range for the class:
(tallest male) - (shortest female) = 56-1 = 55 centimeters.

Case 2: shortest female = 10 centimeters, shortest male = 25 centimeters
Since the range of the male students = 40, tallest male = 25+40 = 65 centimeters.
Resulting range for the class:
(tallest male) - (shortest female) = 65-10 = 55 centimeters.

Since the range for the whole class is THE SAME in each case, SUFFICIENT.

The correct answer is B.
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by ceilidh.erickson » Mon Nov 28, 2016 9:38 am
Since this question asks about RANGE, every other statistic will be irrelevant. Consider these two sets:
[1, 1, 1, 1, 20]
[1, 5, 10, 15, 20]
Both have the same range, but different means, medians, modes, and standard deviations. The only things that have any bearing on the range are the LOWEST and HIGHEST terms in the set.

So, we can easily eliminate statement (1). Information about averages will never help us to answer a RANGE question. (Unless there was additional information about terms being consecutive, and we knew the number of terms. But averages by themselves will not help). We can eliminate answer choices A, D, and even C: this information won't help at all, even if we add it to other information.

All we have to do is determine whether the answer is B or E.

Statement (2) gives us a relationship between the LOWEST terms in each set. Since we already know the range of each set, we can easily represent these values algebraically. Let F be the shortest female student, and M be the shortest male student.

range of female students: F______________________(F+30)
range of male students: M____________________________(M+40)

Since we know that the shortest female student is 15 shorter than M, substitute:

range of female students: (M-15)______________________(F+30)
range of male students: M____________________________(M+40)

range of whole class: (M-15)_____________________________________(M+40)

The difference between these two values is the range: (M+40) - (M-15) = 55.

Statement 2 is sufficient. The answer is B.
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by Mo2men » Mon Nov 28, 2016 2:34 pm
ceilidh.erickson wrote:
So, we can easily eliminate statement (1). Information about averages will never help us to answer a RANGE question. (Unless there was additional information about terms being consecutive, and we knew the number of terms. But averages by themselves will not help). We can eliminate answer choices A, D, and even C: this information won't help at all, even if we add it to other information.
Thanks a lot

Really great visual representation of the problem that makes it easier to solve.

In the above you mentioned that mean won't help to solve for ranges problems. So I have general question based on the above note.

Can you please give some notes about relations that could be relevant to solve stats question? what stats criteria that could help solve stats DS questions?

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by ceilidh.erickson » Mon Nov 28, 2016 2:55 pm
Mo2men wrote:
ceilidh.erickson wrote:
So, we can easily eliminate statement (1). Information about averages will never help us to answer a RANGE question. (Unless there was additional information about terms being consecutive, and we knew the number of terms. But averages by themselves will not help). We can eliminate answer choices A, D, and even C: this information won't help at all, even if we add it to other information.
Thanks a lot

Really great visual representation of the problem that makes it easier to solve.

In the above you mentioned that mean won't help to solve for ranges problems. So I have general question based on the above note.

Can you please give some notes about relations that could be relevant to solve stats question? what stats criteria that could help solve stats DS questions?
My pleasure!

Ok, this may not be a complete list, but here's what comes to mind...

If the question is asking about:

1) MEAN
- helpful: sum, number of terms
- not helpful: median (unless it's an evenly spaced set), range (or smallest or largest value), standard deviation

2) MEDIAN
- helpful: number of terms
- maybe helpful: whether it's an odd or even number of terms
- not helpful: mean (unless evenly spaced), range (or smallest or largest value), standard deviation

3) STANDARD DEVIATION
- helpful: knowing the value of every term
- maybe helpful: range
- not helpful: anything else

4) RANGE
- helpful: smallest and largest values
- not helpful: anything else (unless evenly spaced and you know the # of terms)

5) SUM
- helpful: average and number of terms
- maybe helpful: starting or ending value (if you have # of terms)
- not helpful: median (unless evenly spaced and you know # of terms), range, standard deviation

As a very broad general rule, one statistic alone usually won't tell you much about any other statistics!
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by Mo2men » Mon Nov 28, 2016 3:06 pm
ceilidh.erickson wrote: My pleasure!

Ok, this may not be a complete list, but here's what comes to mind...

If the question is asking about:

1) MEAN
- helpful: sum, number of terms
- not helpful: median (unless it's an evenly spaced set), range (or smallest or largest value), standard deviation

2) MEDIAN
- helpful: number of terms
- maybe helpful: whether it's an odd or even number of terms
- not helpful: mean (unless evenly spaced), range (or smallest or largest value), standard deviation

3) STANDARD DEVIATION
- helpful: knowing the value of every term
- maybe helpful: range
- not helpful: anything else

4) RANGE
- helpful: smallest and largest values
- not helpful: anything else (unless evenly spaced and you know the # of terms)

5) SUM
- helpful: average and number of terms
- maybe helpful: starting or ending value (if you have # of terms)
- not helpful: median (unless evenly spaced and you know # of terms), range, standard deviation

As a very broad general rule, one statistic alone usually won't tell you much about any other statistics!
Again thanks for your great help and prompt response.