SOLUTION: If the standard deviation of 3 numbers (a), (2a+1）and (2) is sqr(1/2), find the value of a. Any help appreciated!

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Question 1037849: If the standard deviation of 3 numbers (a), (2a+1）and (2) is sqr(1/2), find the value of a.
Any help appreciated!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Here is a general outline of how to find the sample standard deviation

Step 1) Add up all the values
Step 2) Divide the sum in step 1 by three (there are three values added up). This is the mean M
Step 3) Subtract each value from the mean M
Step 4) Square each result in step 3
Step 5) Add up the reslts in step 4
Step 6) Divide the result in step 5 by n-1 = 3-1 = 2
Step 7) Take the square root of the result of step 6
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Let's go through all the steps shown in the outline above

---------------------------------------------
Step 1)
(value1)+(value2)+(value3) = (a)+(2a+1)+(2)
(value1)+(value2)+(value3) = 3a+3
The result of step 1 is 3a+3
---------------------------------------------
Step 2)

We divide the result from step 1 (which was 3a+3) by 3 because there are 3 values

(result in step 1)/3 = (3a+3)/3 = a+1

So the mean M is a+1

M = a+1

We will use this in step 3
---------------------------------------------
Step 3)

Subtract each value from the mean M

(value1) - (mean) = (a) - (a+1) = a-a-1 = -1
(value2) - (mean) = (2a+1) - (a+1) = 2a+1-a-1 = a
(value3) - (mean) = (2) - (a+1) = 2-a-1 = 1-a

The results after the subtractions are:
-1
a
1-a
---------------------------------------------
Step 4)

The results of step 3 were -1, a, and 1-a. Let's square the results

(previous result1)^2 = (-1)^2 = 1
(previous result2)^2 = (a)^2 = a^2
(previous result3)^2 = (1-a)^2 = 1-2a+a^2

The results of step 4 are:
1
a^2
1-2a+a^2
---------------------------------------------
Step 5)
Now let's add up the results of the previous step (step 4)

(previous result1)+(previous result2)+(previous result3) = (1)+(a^2)+(1-2a+a^2)
(previous result1)+(previous result2)+(previous result3) = 2a^2-2a+2
---------------------------------------------
Step 6) Divide the previous result by 2 (n-1 = 3-1 = 2)

(previous result)/2 = (2a^2-2a+2)/2 = a^2-a+1
---------------------------------------------
Step 7)

Take the square root of the previous result and we get

sqrt%28a%5E2-a%2B1%29
========================================================================================================
========================================================================================================

That is a lot of work, but we found the sample standard deviation of the three values to be sqrt%28a%5E2-a%2B1%29

Set this equal to sqrt%281%2F2%29 and solve for 'a'


sqrt%28a%5E2-a%2B1%29=sqrt%281%2F2%29
%28sqrt%28a%5E2-a%2B1%29%29%5E2=%28sqrt%281%2F2%29%29%5E2 Square both sides
a%5E2-a%2B1=1%2F2
2a%5E2-2a%2B2=1
2a%5E2-2a%2B2-1=0
2a%5E2-2a%2B1=0

Then I run into a problem: if I use the quadratic formula to solve for 'a', I get two non-real answers. So it's making me think that there is a typo somewhere in the problem. You'll have to ask your teacher about it.


Edit: I redid the problem but instead of dividing by n-1 = 2 (in step 6), I divided by n = 3. This is to reflect the population standard deviation and not the sample standard deviation. Every other step is the same. Doing that leads to sqrt%28%282%2B2a%5E2-2a%29%2F3%29+=+sqrt%281%2F2%29 which leads to a+=+1%2F2. However, I'm not sure which form of standard deviation your teacher wants.