<p>Let X be a group of number {a,b,d,e} and Y be a group of number {a+2,b+2,c+2,d+2,e+2} where a<b<c<d<e.

Which of the following must be true?

I. The median of X is smaller than that of Y.

II. The range of X and the range of Y are the same.

III. The standard deviation of X is greater than that of Y.</p>

<p>II. Since a<b<c<d and each of it had +2, the difference stays the same</p>

<p>The problem lies in III.

I, II are quite obvious...</p>

<p>Bump............</p>

<p>I assume X should be {a,b,c,d,e}.</p>

<p>Let's plug in some numbers:</p>

<p>a=1, b=2, c=3, d=4, e=5</p>

<p>Then X={1,2,3,4,5} and Y={3,4,5,6,7}</p>

<p>The median of X is 3 and the median of Y is 5.

The range of X and Y are both 4.

The standard deviation of X and Y are both approximately 1.14 (just use the stat function on your calculator here)</p>

<p>So I certainly doesn't have to be true, and it would be a very good guess that II and III do have to be true.</p>

<p>Now, a rigorous explanation:

(I) when you increase every number in a set by 2, the median increases by 2 as well.

(II) If (b+2) - (a+2) = b+2-a-2 = b-a. So the ranges are the same.

(III) when you increase every number in a set by 2, the mean increases by 2, and every number in the new set is the same distance from the new mean, as the old numbers were from the old mean . Thus the satndard deviations are the same.</p>

<p>DrSteve's method is correct, but if the standard deviation of X and Y are approximately the same, then "III. The standard deviation of X is greater than that of Y." cannot be correct.</p>

<p>Another way:</p>

<p>I. Because the numbers are different, the median has to be different.

II. While the numbers are different, In Y they each go up the same amount, so the range stays the same.

III. Likewise, because they increase at the same value, the stand dev stays the same.</p>

<p>So, II.</p>

<p>Craig Gonzales</p>

<p>Thanks for the correction Craig. I thought that number III said "equal" instead of "greater."</p>

<p>About your method: your explanation for number I is flawed. Here's a counterexample:</p>

<p>The sets {2,4} and {1,5} both have different numbers, but they have the same median of 3. </p>

<p>This situation doesn't happen in this problem because whenever you add the **same** number to each number in a set, that number is added to the median as well.</p>

<p>I assume X should be {a,b,c,d,e}.

No, it's really {a,b,d,e}</p>

<p>And the answer is that, only (I) is incorrect. It bothers me.</p>

<p>I understand what will happen to the standard deviation if we increase each set of number by 2. The differences between each number and the mean stay the same and hence the sum of their squares stay the same. So standard deviation should stay the same for constant no. of terms.

But for {a,b,d,e}, it's completely another case.</p>

<p>But even comparing the sets {a,b,d,e} and {a=2, b+2, c+ 2, d+2, e+2}, the two sets would have the same range. You would not be able to make a determination about the standard deviation though. </p>

<p>So I don't see how the answer is I only...</p>

<p>And of course, this is NOT an SAT question. So if you are studying this under the "prepare for harder stuff so that the SAT will seem easier" plan, I think you are wasting your time. In addition, I bet the missing 'c' is in fact a typo -- one that almost creates a valid, difficult question, but not the one intended. </p>

<p>Where is this from?</p>

<p>The answer is II, III, and I is incorrect...

The determination of standard deviation is one which matters me most. I have no problem for the other two.

The question's from my local math exam (I'm an international), so it's not specifically for SAT. It's unlikely a typo, as this exam was released by the government's exam authority and distributed to most of the senior students of the city.</p>

<p><a href="http://img13.imageshack.us/img13/8379/72527863.jpg%5B/url%5D">http://img13.imageshack.us/img13/8379/72527863.jpg</a>

It is exactly how the question looks like. I just changed some symbols for convenience.

The answer is C.</p>

<p>OK, I think they are playing with this idea: Standard deviation is like an average of how far each data point is from the average. (Technically, you square the differences, add em up, divide by how many and then take the square root). By adding another number to the list that is forced to be "in the middle" of the range, you improve the standard deviation. Best way to see this: make up data sets. Try 0, 1, 99, 100. Calculate the std dev. Then introduce another number between 1 and 99. It will be closer to the mean than the other numbers and when you recalculate the std dev, it will be lower.</p>

<p>(But I admit I have not tried this yet and part of me suspects that there could be a counterexample...if anyone has one, let's see it.)</p>