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If

 \sum_{i=1}^{9}    {\left ( x_{i} - 5 \right )} =9 

and  \sum_{i=1}^{9}  {\left ( x_{i} - 5 \right )^{2}} =45, 

then the standard deviation of the 9 items x1 , x2, ........x9 is : 

  • Option 1)

    3

  • Option 2)

    9

  • Option 3)

    4

  • Option 4)

    2

 

Answers (1)

best_answer

\sum_{i=1}^{9}x_{i}= \sum_{i=1}^{9}(5)=9

\sum_{i=1}^{9}x_{i}- 45= 9\Rightarrow \sum_{i=1}^{9}x_{i}=54

\sum_{i=1}^{9}x_{i^{2}}-10\sum_{i=1}^{9}x_{i}+\sum_{i=1}^{9}25= 45

\sum_{i=1}^{9}x_{i}^{2}=360

 

 

 

 

Variance = \frac{\sum _{i}^{2}}{9}- (\frac{\sum x_{i}}{9})^{2}= 4

 

S.D = 2

 

Standard Deviation -

If x1, x2...xn are n observations then square root of the arithmetic mean of 

\sigma = \sqrt{\frac{\sum \left ( x_{i}-\bar{x} \right )^{2}}{n}}

\bar{}

- wherein

where \bar{x} is mean

 

 

Variance -

In case of discrete data 

\dpi{100} \sigma ^{2}= \left ( \frac{\sum x_{i}^{2}}{n} \right )-\left ( \frac{\sum x_{i}}{n} \right )^{2}

-

 

 


Option 1)

3

Option 2)

9

Option 3)

4

Option 4)

2

Posted by

Himanshu

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