Let a, b, c, d, e, be the observations with m and standard deviation s. The standard deviation of the observations a+k, b+k, c+k, d+k, e+k is `s` (b) `k s` (c) `s+k` (d) `s/k`
A. ks
B. 2
C. s+k
D. `(s)/(k)`

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Correct Answer - B
Given observations are a, b, c, d and e.
Mean `=m=(a+b+c+d+e)/(5)`
`sum x_(i)=a+b+c+d+e=5 m`
New mean `=(a+k+b+k+c+k+d+k+e+k)/(5)`
`=((a+b+c+d+e)+5k)/(5)=m+k`
`therefore S.D. =sqrt((sum(x_(i)^(2)+k^(2)+2kx_(i)))/(n)-(m^(2)+k^(2)+2mk))`
`=sqrt((sum x_(i)^(2))/(n)-m^(2)+(2k sum x_(i))/(n)-2mk)`
`=sqrt((sum x_(i)^(2))/(n)-m^(2)+2km-2mk)" " [because (sum x_(i))/(n)=m]`
`=sqrt((sum x_(i)^(2))/(n)-m^(2))`
=s

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