## Tuesday, October 25, 2011

### this week's MGRE Math Beast Challenge problem

MGRE's newest Math Beast Challenge problem is here:

This Week's Problem: "Likely Story"

Q:

{8, 10, 11, 16, 20, 22, 25, x}

In the set above, x is an odd integer between 13 and 21, inclusive. Each possible x value is equally probable.

Which of the following statements has the highest probability of being true?

A:

(A) x is in the second quartile of the set
(B) x is in the third quartile of the set
(C) x differs from the median of the set by less than 2
(D) x differs from the median of the set by more than 2
(E) x is less than the range of the set

This feels more like one of the GRE's stats problems. Remember quartiles? At their most basic, quartiles are three numbers that divide a set of numbers into four equal "quarters." To find those four quarters, which are spaced evenly throughout the data: first determine the median by arranging the numbers and finding the middle term. Then, to find the lower quartile, find the median of the range defined by the first median and the lowest data point. To find the upper quartile, find the median of the range defined by the first median and the highest number in the set. If a set has an odd number of elements, the median is the middle value. If the set has an even number of elements, the median is the average of the two numbers that comprise the middle of the set.

Oh, and did I get last week's problem correct? I said the answer was once again D. MGRE says...

No.

Their solution:

The answer is B: Quantity B is greater.

We could simplify both quantities, which are both positive by definition, then compare individual terms:

Quantity A: (y!)/[5(y-2)!] = [y(y-1)]/5

Quantity B: (y+1)!/[3(y-1)!)] = [y(y+1)]/3

The factor of y is common to both Quantities.

(y – 1) in Quantity A < (y + 1) in Quantity B. Thus, the numerator of Quantity A is smaller than that of Quantity B. 5 > 3, so the denominator of Quantity A is greater than that of Quantity B.

Therefore, Quantity A < Quantity B.

Damn, damn, damn. I suspect I did indeed misread the problem, as I noted in my second comment to it.

_

1. From the information given in the problem, we know that x can be any of five values: 13, 15, 17, 19, 21. A median is a middle value in a set of numbers; if the number of terms in the set is odd, then the median is the middle term. If the number of terms is even, then the median is the average of the two middle terms.

So let's plot this out, with all the possibilities:

x = 13
{8, 10, 11, 13, 16, 20, 22, 25}
1st Quartile = 10.5
Median = 14.5
3rd Quartile = 21

x = 15
{8, 10, 11, 15, 16, 20, 22, 25}
1st Quartile = 10.5
Median = 15.5
3rd Quartile = 21

x = 17
{8, 10, 11, 16, 17, 20, 22, 25}
1st Quartile = 10.5
Median = 16.5
3rd Quartile = 21

x = 19
{8, 10, 11, 16, 19, 20, 22, 25}
1st Quartile = 10.5
Median = 17.5
3rd Quartile = 21

x = 21
{8, 10, 11, 16, 20, 21, 22, 25}
1st Quartile = 10.5
Median = 18
3rd Quartile = 21.5

What are the chances that x lies in the first quartile? Zero. In fact, that answer's not even listed as a possibility, so we move on.

What are the chances that x lies in the second quartile range, i.e., between the first quartile and the median? The answer is 2/5.

What about x's being in the third quartile range, i.e., between the median and the third quartile? The answer is 3/5.

What about x's differing from the median by less than 2? The answer is 4/5.

What about x's differing from the median by more than 2? The answer is 1/5.

What about x being less than the range of the set? The range is defined as the difference between the largest and smallest terms of the set-- in this case, 25 and 8. Well 25 - 8 = 17, so the range is 17. In how many cases can x be less than 17? The answer is 2/5, because only 13 and 15 are less than 17, out of the five possible values of x.

It seems to me that the correct answer is C, since there's a 4 out of 5 chance that x will differ from a given median by less than 2.

Do you concur?

2. In case it's not obvious from the above, I determined the first and third quartiles by averaging the second and third terms, and the sixth and seventh terms, respectively.