1.

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. \(\frac{s}{k}\)  

Answer»

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


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