Sigma(x_{i} - bar{x}) = ldots ldots ldots | vedclass

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(D) The mean of a set of observations is defined as $\bar{x} = \frac{\Sigma x_{i}}{n}$,where $n$ is the number of observations.Therefore,$\Sigma x_{i}

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$0$

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(D) The mean of a set of observations is defined as $\bar{x} = \frac{\Sigma x_{i}}{n}$,where $n$ is the number of observations.Therefore,$\Sigma x_{i} = n\bar{x}$.Now,consider the expression $\Sigma(x_{i} - \bar{x})$.This can be expanded as $\Sigma x_{i} - \Sigma \bar{x}$.Since $\bar{x}$ is a constant,$\Sigma \bar{x} = n\bar{x}$.Substituting the values,we get $\Sigma x_{i} - n\bar{x} = n\bar{x} - n\bar{x} = 0$.Thus,the sum of deviations of observations from their mean is always $0$.

Marks	$30-35$	$35-40$	$40-45$	$45-50$	$50-55$	$55-60$	$60-65$
Frequency	$14$	$16$	$18$	$23$	$18$	$8$	$3$

Class	$5-10$	$10-15$	$15-20$	$20-25$	$25-30$	$30-35$
Frequency	$11$	$20$	$35$	$20$	$8$	$6$

Class	$65-85$	$85-105$	$105-125$	$125-145$	$145-165$	$165-185$	$185-205$
Frequency	$4$	$5$	$13$	$20$	$14$	$7$	$4$

$\Sigma(x_{i} - \bar{x}) = \ldots \ldots \ldots$

Similar Questions

Find the median of the following frequency distribution:

Class $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ $55-60$ $60-65$ $65-70$ $70-75$

Frequency $1$ $2$ $5$ $10$ $17$ $15$ $9$ $5$ $3$

The percentage of marks obtained by $100$ students in an examination are given below:

Marks $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ $55-60$ $60-65$

Frequency $14$ $16$ $18$ $23$ $18$ $8$ $3$

Determine the median percentage of marks.

Find the mean,median and mode of the following frequency distribution:

Class $5-10$ $10-15$ $15-20$ $20-25$ $25-30$ $30-35$

Frequency $11$ $20$ $35$ $20$ $8$ $6$

Consider the data:

Class $65-85$ $85-105$ $105-125$ $125-145$ $145-165$ $165-185$ $185-205$

Frequency $4$ $5$ $13$ $20$ $14$ $7$ $4$

The difference between the upper limit of the median class and the lower limit of the modal class is:

For a given frequency distribution,$n=100, A=20$ and $\bar{x}=20$. Then,$\Sigma f_{i} d_{i}=\ldots \ldots \ldots . .$

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