If `x_(i)'s` are the mid-points of the class intervals of grouped data, `f_(i)'s` are the corresponding frequencies and `bar(x)` is the mean, then `sum(f_(i)x_(i)-bar(x))` equal to

If x_(i) ' s are the midpoints of the class intervals of grouped data, f_(i) 's are the corresponding frequencies and barx is the mean, then what is Sigma (f_(i)x_(i)-barx) equal to ?

If x_(i) 's are the mid-points of the class intervals of grouped data f_(i)'s are the corresponding frequencies and x is the mea, then sum(f_(i)x_(i)-x) is equal to

The average of n numbers x_(1), x_(2), ……. X_(n) is bar(x) . Then the value of sum_(i=1)^(n) (x_(1) - bar(x)) is equal to

Find the mean (bar(X)) if sumf_(i)x_(i)=75andsumf_(i)=15 .

In the formula barx=a+(sum f_(i)d_(i))/(sum f_(i)) for finding the mean of grouped data, ' d_(i) 's are deviations from the

The measurments ( in mm) of the diameters of the head of the screws are given below : Calculate the mean diameter of the screw by ' Assumed Mean Method' . Steps to calculate mean by this method : (1) Assume mean A. In this example , A = 40. (Generally, the class mark of the class having the maximum frequency or the class mark of the middle class is chosen as the assumed mean.) (2) In column 1, write the class intervals,(Here, diameter). (3) In column 2, write the corresponding class marks. (4) In column 3, write the value of d_(i), " where " d_(i) = x_(i) - A . A is the assumed mean. (5) In column 4, write the frequencies (f_(i)) . Write Sigma f_(i) at the bottom of this column. (6) In column 5, write the product (f_(i) xx d_(i)) and their sum Sigma f_(i) d_(i) . Calculate bar d " and " bar X using the following formulae : bard = (Sigma f_(i)d_(i))/(Sigma f_(i))." Mean" barX = A + bar d . [d : mean of deviations.]

If bar(X) is the mean of a distribution of X, then under usual notation sum sum_(i=1)^(n)f_(i)(x_(2)-bar(x)) is

In the formula bar(x)=a+(sumf_(i)d_(i))/(sumf_(i)) for finding the mean of grouped data d_(i) 'S and deviation from a of