Asked by Ms Fatimah
If x is positive,and the standard deviation of x -1,x + 1,3,2x - 1 and x + 3 is 2√2, find the value of x and hence calculate the mean deviation.?
Answers
Answered by
Reiny
Mean = (x -1 + x + 1 + 3 + 2x - 1 + x + 3)/5 = (5x + 5)/5 = x+1
then (x-1 - x-1)^2 = 4
(x+1 - x-1)^2 = 0
(3 - x-1)^2 = 4 - 4x + x^2
(2x-1 - x-1)^2 = x^2 - 4x + 4
(x+3 - x-1)^2 = 4
SD = √[(4 + 0 + 4-4x+x^2 + 4-4x+x^2 + 4)/5]
= √[(16 -8x + 2x^2)/5 = 2√2
square both sides and simplify
2x^2 - 8x + 16 = 40
x^2 - 4x - 12 = 0
(x-6)(x+2) = 0
x = 6 or x = -2
but x> 0, so
x = 6
then (x-1 - x-1)^2 = 4
(x+1 - x-1)^2 = 0
(3 - x-1)^2 = 4 - 4x + x^2
(2x-1 - x-1)^2 = x^2 - 4x + 4
(x+3 - x-1)^2 = 4
SD = √[(4 + 0 + 4-4x+x^2 + 4-4x+x^2 + 4)/5]
= √[(16 -8x + 2x^2)/5 = 2√2
square both sides and simplify
2x^2 - 8x + 16 = 40
x^2 - 4x - 12 = 0
(x-6)(x+2) = 0
x = 6 or x = -2
but x> 0, so
x = 6
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