Question 1037849
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


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Step 1) 

(value1)+(value2)+(value3) = (a)+(2a+1)+(2) 
(value1)+(value2)+(value3) = 3a+3

The result of step 1 is 3a+3

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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

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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

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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

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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

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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

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Step 7) 


Take the square root of the previous result and we get


{{{sqrt(a^2-a+1)}}}

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That is a lot of work, but we found the sample standard deviation of the three values to be {{{sqrt(a^2-a+1)}}}


Set this equal to {{{sqrt(1/2)}}} and solve for 'a'



{{{sqrt(a^2-a+1)=sqrt(1/2)}}}
{{{(sqrt(a^2-a+1))^2=(sqrt(1/2))^2}}} Square both sides
{{{a^2-a+1=1/2}}}
{{{2a^2-2a+2=1}}}
{{{2a^2-2a+2-1=0}}}
{{{2a^2-2a+1=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((2+2a^2-2a)/3) = sqrt(1/2)}}} which leads to {{{a = 1/2}}}. However, I'm not sure which form of standard deviation your teacher wants.