If $x_{i}$ are the midpoints of the class intervals of grouped data,$f_{i}$ are the corresponding frequencies,and $\bar{x}$ is the mean,then $\sum (f_{i} x_{i} - f_{i} \bar{x})$ is equal to

  • A
    $0$
  • B
    $-1$
  • C
    $1$
  • D
    $2$

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Similar Questions

$\sum_{i=1}^{10} (x_i - \bar{x}) = \dots$

The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories $(U.T.)$ of India. Find the mean percentage of female teachers by all the three methods discussed in this section.
Percentage of female teachers$15$-$25$$25$-$35$$35$-$45$$45$-$55$$55$-$65$$65$-$75$$75$-$85$
Number of States/$U$.$T$.$6$$11$$7$$4$$4$$2$$1$

For a given frequency distribution,$n=100$ and $\Sigma f_{i} x_{i}=245$. Then,$\bar{x}=\ldots \ldots \ldots \ldots$

$A$ class teacher has the following absentee record of $40$ students of a class for the whole term. Find the mean number of $days$ a student was absent.
$\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Number of days} & 0-6 & 6-10 & 10-14 & 14-20 & 20-28 & 28-38 & 38-40 \\ \hline \text{Number of students} & 11 & 10 & 7 & 4 & 4 & 3 & 1 \\ \hline \end{array}$

In the formula $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$,$d_{i} =$ .........

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