MHB How Do I Calculate Standard Deviation with Changed Values in a Set?

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To calculate the change in standard deviation when altering the set {11, 12, 13, 14, 15} by decreasing the minimum value by 2 and increasing the maximum by 2, the initial standard deviation is found to be approximately 1.414. After the changes, the new set becomes {9, 12, 13, 14, 17}, resulting in a new standard deviation of approximately 2.608. The increase in standard deviation is calculated as approximately 1.914. The mean remains unchanged at 13 due to the symmetric nature of the alterations. Understanding these calculations is essential for GRE preparation and statistics comprehension.
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Can someone guide me through the process of solving the following? Many thanks.

"Given the set {11, 12, 13, 14, 15}, by approximately how much does the standard deviation change if the least value is decreased by 2 and the greatest value is increased by 2?"

These questions are all over the GRE, and I'm lost.
 
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Super Rough (but sometimes useful) approximation for Standard deviation...Range/6 = (max-min)/6

Indirectly...
As is: (15-11)/6 = 2/3
Altered: (17-9)/6 = 4/3
4/3 - 2/3 = +2/3

Directly...
+4 / 6 = +2/3
 
If an approximation is not good enough, the definition of "standard deviation" is "The square root of the mean of the difference from the mean, squared". That is $\sqrt{\frac{\sum (x- \mu)^2}{n}}$ where $\mu$ is the mean and n is the number of terms. (Who ever gave you this problem probably expected you to know that!)

Here, the data set is {11, 12, 13, 14, 15}. The mean is $\frac{11+ 12+ 13+ 14+ 15}{5}= \frac{65}{5}= 13$. Notice that this is the "middle" number in the set. That works because this is an "arithmetic sequence".

Now, subtract that mean from each number and square:
11- 13= -2 and squaring, 4.
12- 13= -1 and squaring, 1.
13- 13= 0 and squaring, 0.
14- 13= 1 and squaring, 1.
15- 13= 2 and squaring, 4.

The mean of those numbers is $\frac{4+ 1+ 1+ 4}{5}= \frac{10}{5}= 2$.

The standard deviation is $\sqrt{2}$.

Now do the same with the altered data set, {9, 12, 13, 14, 17}, decreasing the smallest number, 11, by 2 to get 9 and increasing the largest number, 15, by 2 to get 17. The mean is $\frac{9+ 12+ 13+ 14+ 17}{5}= \frac{65}{5}= 13$ again. (Think about why that is true.)

Now subtract that mean from each number and square:
9- 13= -4 and squaring, 16.
12- 13= -1 and squaring, 1.
13- 13= 0 and squaring, 0.
14- 13= 1 and squaring, 1.
17- 13= 4 and squaring, 16.

The mean of those numbers is $\frac{16+ 1+ 0+ 1+ 16}{5}= \frac{34}{5}= 6.8$.

The standard deviation Is $\sqrt{6.8}= \sqrt{4(1.7)}= 2\sqrt{1.7}$.

The initial standard deviation was $\sqrt{2}$ which is approximately 1.414 while the new standard deviation is $\sqrt{6.8}$ which is approximately 2.608. The standard deviation has increased by approximately 2.608- 1.414= 1.914
 
HallsofIvy said:
1.914

And THIS is why we called it "Super Rough". :-)
 
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