For a frequency distribution,the standard deviation is computed by:
- A$\sigma = \frac{\sum f(x - \bar{x})}{\sum f}$
- B$\sigma = \frac{\sqrt{\sum f(x - \bar{x})^2}}{\sum f}$
- C$\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}}$
- D$\sigma = \sqrt{\frac{\sum f(x - \bar{x})}{\sum f}}$
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Similar Questions
The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where $A$ is a positive integer,has a variance of $160$. Determine the value of $A$.
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where $A$ is a positive integer,has a variance of $160$. Determine the value of $A$.
Medium
View SolutionThe mean of two samples of size $200$ and $300$ were found to be $25$ and $10$ respectively. Their standard deviations $(S.D.)$ are $3$ and $4$ respectively. Then,the variance of the combined sample of size $500$ is:
The variance of the observations $112, 116, 120, 125, 132$ is $......$
Medium
View SolutionIf $1$ is added to each of the first $10$ natural numbers,then the variance of the numbers so obtained is:
Variance of first $n$ natural numbers is $\qquad$ .
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