MD=SD=R/2 (Sign are natural)
Without going to the R(Range)2\dfrac{R\text{(Range)}}{2}2R(Range) part,
Why do you think MD=SDMD = SDMD=SD ?
That is NOT generally true (although data can be constructed such that MD = SD).
For example: x=[−10,0,10],SD(x)=(102+0+102)3= 2.72x = {[-10, 0, 10]}, SD(x) = \sqrt{\dfrac{(10^2 + 0 + 10^2)}{3}} = ~2.72x=[−10,0,10],SD(x)=3(102+0+102)= 2.72
(or another answer if you divide by N−1N-1N−1 instead of NNN )
MD(x)=(10+0+10)3=6.66...MD(x) = \dfrac{(10 + 0 + 10)}{3} = 6.66...MD(x)=3(10+0+10)=6.66...
SD(x)≠MD(x)SD(x) \neq MD(x)SD(x)=MD(x)
Hence, MD≠SD≠R2MD\neq SD\neq \dfrac{R}{2}MD=SD=2R
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