Show that for any 2 numbers a and b, standard deviation is given by |a-b|÷2

Question

The variance is taken to solve it and to show the s²= |a+b|

                 2

 

I don't know why

               

Answer 1

Show that for any 2 numbers a and b, standard deviation is given by |a-b|÷2

Note that standard deviation is always positive or zero.

 

1.  Take the mean of the numbers:  (a+b)/2

 

2.  Subtract the mean from each of the original numbers and square the differences:

 

    a - (a+b)/2 = (2a - a - b)/2 = (a-b)/2

    [(a-b)/2]² = (a²-2ab+b²)/4

 

    b - (a+b)/2 = (2b-a-b)/2 = (b-a)/2

    [(b-a)/2]² = (b²-2ab + a²)/4

 

3.  Take the mean of the squares above

    [(a²-2ab+b²)/4 + (b_²-2ab+a²)/4]/2

     = (2a² - 4ab + 2b²)/8

     = 2(a² - 2ab + b²)/8

     = (a²- 2ab + b²)/4

     = (a-b)²/4

    

4.  Take the square root of the above

     √[(a-b)²/4] = ±(√(a-b)²)/2 = ±(a-b)/2

     Since sd is always positive or zero

     ±(a-b)/2 = |a-b|/2