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

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

Choose the correct answer from the given four options: In the formula x=a+(sumx_(1)d_(i))/(sumf_(i)) for finding the mean of grouped data d_(i) 's are the deviations from a of

In the formula barx=a+((sumf_iu_i)/(sumf_i))xx h, u_i=

In the formula barx=a+((sumf_iu_i)/(sumf_i))xx h, u_i=

In the formula bar(x)=a+h(sumf_(i)u_(i))/(sumf_(i)) for finding the mean of grouped frequency distribution u_(i) is equal to

In the formula bar(x)=a+h(sumf_(i)u_(i))/(sumf_(i)) for finding the mean of grouped frequency distribution u_(i) is equal to

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

In Calculating the mean of grouped data, grouped in classes of equal width, we may use the formula bar(x)=a+(sumf_(i)d_(i))/(sumf_(i)) Where, a is the assumed mean, a must be one of the mid point of the classes. Is the last statement correct? Justify your answer.

In Calculating the mean of grouped data, grouped in classes of equal width, we may use the formula bar(x)=a+(sumf_(i)d_(i))/(sumf_(i)) Where, a is the assumed mean, a must be one of the mid point of the classes. Is the last statement correct? Justify your answer.

barx=A+(sumf_i d_i)/(sumf_i) is the formula to find mean for an ungrouped data. In this formula letter A represents for