Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is
A. s
B. ks
C. s + k
D. \(\frac{s}{k}\)
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Let a, b, c, d, e be the observation and mean is m
m = \(\frac{a+b+c+d+e}{5}\)
Let suppose new mean be m1
m1 = \(\frac{a+k+b+k+c+k+d+k+e+k}{5}\)
m1 = \(\frac{5k}{5}+\frac{a+b+c+d+e}{5}\)
M1 = m + k
Now, The standard deviation
S = \(\sqrt{\frac{(a-m)^2+(b-m)^2+(c-m)^2+(e-m)^2}{5}}\)
So, The standard deviation for new observation
S1 =
\(\sqrt{\frac{(a+k-n)^2+(b+k-n)^2+(c+k-n)^2+(e+k-n)^2+(d+k-n)^2}{5}}\)
Now, we can compare both observation
a+k-n=a+k-(m+k)
a+k–n = a+k–m- k
a+k-n=a-m
Similary
b+k-n=b-m
c+k-n=c-m
d+k-n=d-m
e+k-n=e-m
when we substitute the values, we get,
S1 = S
Hence, The Sd is S