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  • 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

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}

Answers (1)

Given observations are a, b, c, d, e

So, the mean of the 5 observations is

\\ m=\frac{a+b+c+d+e}{5}=\frac{ \sum x_{i}}{5} \\\sum x_{i}=5m=a+b+c+d+e \\

And the standard deviation of the 5 observations is

\\ \sigma =\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}- \left( \frac{ \sum x_{i}}{5} \right) ^{2}}=\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2}} \\

Now we will find the mean and standard deviation of the observations a+k , b+k, c+k, d+k, e+k  we get

So, the mean of these 5 observations is

\\ m_{1}=\frac{ \left( a+k \right) + \left( b+k \right) + \left( c+k \right) + \left( d+k \right) + \left( e+k \right) }{5} \\ \\ =\frac{a+b+c+d+e}{5}+k=m+k \\ \\ \sigma _{1}=\sqrt {\frac{ \sum \left( x_{i}+k \right) ^{2}~}{5}- \left( \frac{ \sum \left( x_{i}+k \right) }{5} \right) ^{2}}=\sqrt {\frac{ \sum \left( x_{i}^{2}+k^{2}+2kx_{i} \right) }{5}- \left( m+k \right) ^{2}} \\ \\ =\sqrt {\frac{ \sum \left( x_{i}^{2} \right) }{5}+\frac{5k^{2}}{5}+\frac{2k \sum \left( x_{i} \right) }{5}- \left( m+k \right) ^{2}} \\ \\ =\sqrt {\frac{ \sum \left( x_{i}^{2} \right) }{5}+k^{2}+2mk-m^{2}-k^{2}-2mk} \\ \\ =\sqrt {\frac{ \sum \left( x_{i}^{2} \right) }{5}-m^{2}} \\ \\ \sigma _{1}= \sigma \\

 

 

 

Posted by

infoexpert21

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