d standard deviation

Hi J_shreyans,

In real basic terms, Standard Deviation is a measure of how "spread out" a group of numbers is. Numbers that are "close together" will create a small Standard Deviation, while numbers that are really "spread out" will create a large Standard Deviation.

Here, we have 3 values: X, Y and Z. We're told that they have a Standard Deviation of D.

To help you visualize what's going on here, let's TEST VALUES…

X = 1
Y = 2
Z = 3
Standard Deviation = D (don't worry about calculating it, the GMAT will NEVER ask you to calculate a Standard Deviation)

Notice the "spread" of the above numbers; we'll come back to this in a moment.

The question then asks us to think about X+5, Y+5 and Z+5. Using the values from above, we would have….
1+5 = 6
2+5 = 7
3+5 = 8

Now, notice how these values have the exact SAME SPREAD as the initial set of numbers? This group of numbers has the SAME STANDARD DEVIATION as the original set of numbers. Thus, the Stand Deviation is still D.

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

For the purposes of the GMAT, it's sufficient to think of Standard Deviation as the Average Distance from the Mean. Here's what I mean:

Consider these two sets: Set A {7,9,10,14} and set B {1,8,13,18}
The mean of set A = 10 and the mean of set B = 10
How do the Standard Deviations compare? Well, since the numbers in set B deviate the more from the mean than do the numbers in set A, we can see that the standard deviation of set B must be greater than the standard deviation of set A.

Alternatively, let's examine the Average Distance from the Mean for each set.

Set A {7,9,10,14}
Mean = 10
7 is a distance of 3 from the mean of 10
9 is a distance of 1 from the mean of 10
10 is a distance of 0 from the mean of 10
14 is a distance of 4 from the mean of 10
So, the average distance from the mean = (3+1+0+4)/4 = 2

B {1,8,13,18}
Mean = 10
1 is a distance of 9 from the mean of 10
8 is a distance of 2 from the mean of 10
13 is a distance of 3 from the mean of 10
18 is a distance of 8 from the mean of 10
So, the average distance from the mean = (9+2+3+8)/4 = 5.5

IMPORTANT: I'm not saying that the Standard Deviation of set A equals 2, and I'm not saying that the Standard Deviation of set B equals 5.5 (They are reasonably close however).

What I am saying is that the average distance from the mean can help us see that the standard deviation of set B must be greater than the standard deviation of set A.
More importantly, the average distance from the mean is a useful way to think of standard deviation. This model is a convenient way to handle most standard deviation questions on the GMAT.
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If we apply this definition to the original question, we can see that adding 5 to each of the values has NO EFFECT on their distance from mean. Therefore, the standard deviation will not change.

Cheers,
Brent

Here are a few more practice questions where we can apply the concept of "average distance from the mean" as an approximation for Standard Deviation:

https://www.beatthegmat.com/standard-dev ... 74384.html
https://www.beatthegmat.com/standard-dev ... 69584.html
https://www.beatthegmat.com/range-and-sd-t89159.html