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

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.

The mean of a discrete frequency distribution x_i//f_i ; i=1,\ 2,\ ddot,\ n is given by (a) (sumf_i x_i)/(sumf_i) (b) 1/n\ sum_(i=1)^nf_i x_i (c) (sumi=1nf_i x_i)/(sumi=1n x_i) (d) (sumi=1nf_i x_i)/(sumi=1n i)

For a frequency distribution standard deviation is computed by applying the formula (a) sigma=sqrt((sumfd^2)/(sumf)-((sumfd)/(sumf))^2) (b) sigma=sqrt(((sumfd^2)/(sumf))-(sumfd^2)/(sumf)) (c) sigma=sqrt((sumfd^2)/(sumf)-(sumfd)/(sumf)) (d) sqrt(((sumfd)/(sumf))^2-(sumfd^2)/(sumf))

For a frequency distribution standard deviation is computed by applying the formula (a) sigma=sqrt((sumfd^2)/(sumf)-((sumfd)/(sumf))^2) (b) sigma=sqrt(((sumfd)/(sumf))^2-(sumfd^2)/(sumf)) (c) sigma=sqrt((sumfd^2)/(sumf)-(sumfd)/(sumf)) (d) sigma=sqrt(((sumfd)/(sumf))^2-(sumfd^2)/(sumf))

Use of the formula (a+b) (a-b) = a^(2) - b^(2) to find the value of : (i) 107 xx 93

Statement-1 : {:(x:,5,6,7,8,9),(y:,4,6,12,-,8):} for given data mean of x(barx) is found to be 7.3 . The missing frequency is 10 Statement-2: In case of a frequency distribution , barx=(Sigmaf_ix_i)/(Sigmax_i)

If u_i=(x_i-25)/(10),sumf_i u_i=20 ,sumf_i=100 , then x = 23 (b) 24 (c) 27 (d) 25

If the mean of a frequency distribution is 8.1 and sumf_i x_i=132+5k ,\ \ sumf_i=20 , then k= (a) 3 (b) 4 (c) 5 (d) 6