nikhilgmat31 wrote:GMATGuruNY wrote:nikhilgmat31 wrote:
Hi Mitch,
Can you please explain statement 1 by picking some numbers. I am not able to visualize any numbers to prove statement 1 is not sufficient.
All you need to know is this:
There is NO direct relationship between RANGE and SD.
A = 0,2,2,2,5
B = 0,0,2,4,4
In this case, the SD of A ≈ 1.7, while the SD of B = 2, with the result that B has the greater SD.
Here, A has the smaller SD because three of the data points -- 2,2,2 -- are close to the mean of 2.2.
A = 0,0,2,5,5
B = 0,0,2,4,4
In this case, the SD of A ≈ 2.5, while the SD of B = 2, with the result that A has the greater SD.
Here, A has the greater SD because four of the data points -- 0,0,5,5 -- are far from the mean of 2.4.
Hi Mitch,
How you calculate SD so quickly and reach at SD value
A = 0,2,2,2,5
B = 0,0,2,4,4
In this case, the SD of A ≈ 1.7, while the SD of B = 2
Please check if below one is correct.
1. calculate mean
2. Find squares of difference of each number from mean
3. Find sum of all the squares
4. Divide the sum by numbers of items in set. - It will give Variance
5. Square root of Variance is SD.
You do NOT need to know how to calculate SD.
Generally, problems about SD can be solved conceptually.
Some key concepts:
1. A small SD means that the data points are clustered close to the mean.
2. A large SD means that the data points are spread far from the mean.
3. If SD=0, then all of the data points are equal.
4. Generally, there is no direct relationship between the MEAN of a set and its SD; a set with a greater mean can have a smaller SD, and vice versa.
5. Generally, there is no direct relationship between the RANGE of a set and its SD; a set with a greater range can have a smaller SD, and vice versa.
6. If every data point increases or decreases by a constant k, then the SD does not change.
7. If every data point increases or decreases by the same PERCENTAGE, then the SD also increases or decreases by that percentage.
A more extreme case to illustrate Concept 5 and prove that Statement 1 is insufficient:
A = {0, 50, 50, 50, 50, 50, 100}
B = {1, 1, 1, 50, 99, 99, 99}
Here:
Both sets have a mean of 50.
Range of A > Range of B.
SD of A < SD of B.
Set A clearly has the SMALLER SD because most of its data points are EQUAL to the mean of 50, while most of the data points in B are FAR from the mean of 50.
A = {0, 0, 0, 50, 100, 100, 100}
B = {1, 50, 50, 50, 50, 50, 99}
Here:
Both sets have a mean of 50.
Range of A > Range of B.
SD of A > SD of B.
Set A clearly has the GREATER SD because most of its data points are FAR from the mean of 50, while most of the data points in B are EQUAL to the mean of 50.
As the case above illustrates, we can prove that statement 1 is insufficient WITHOUT calculating the SD for each set of values.
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