The mean deviation for n observations x1...

The mean deviation for `n` observations `x_1, x_2, ......... , x_n` from their mean ` X ` is given by (a)`sum_(i=1)^n(x_i- X )` (b) `1/nsum_(i=1)^n|x_i- X |` (c) `sum_(i=1)^n(x_i- X )^2` (d) `1/nsum_(i=1)^n(x_i- X )^2`

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The mean deviation for n observations x_1, x_2, ......... , x_n from their mean X is given by (a) sum_(i=1)^n(x_i- X ) (b) 1/nsum_(i=1)^n(x_i- X ) (c) sum_(i=1)^n(x_i- X )^2 (d) 1/nsum_(i=1)^n(x_i- X )^2

Mean deviation for n observation x_(1),x_(2),…..x_(n) from their mean bar x is given by

The mean deviation for n observations x_(1),x_(2)…….x_(n) from their median M is given by (i) sum_(i=1)^(n)(x_(i)-M) (ii) (1)/(n)sum_(i=1)^(n)|x_(i)-M| (iii) (1)/(n)sum_(i=1)^(n)(x_(i)-M)^(2) (iv) (1)/(n)sum_(i=1)^(n)(x_(i)-M)

Let x_1, x_2, ... ,x_n be n observations and barX be their arithmetic mean. The standard deviation is given by (a) sum_(i=1)^n(x_i- barX )^2 (b) 1/nsum_(i=1)^n(x_i- barX )^2 (c) sqrt(1/nsum_(i=1)^n(x_i- X )^2) (c) 1/nsum_(i=1)^n x_i^2- barX ^2

The standard deviation of n observations x_(1), x_(2), x_(3),…x_(n) is 2. If sum_(i=1)^(n) x_(i)^(2) = 100 and sum_(i=1)^(n) x_(i) = 20 show the values of n are 5 or 20

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)

If barx represents the mean of n observations x_(1), x_(2),………., x_(n) , then values of Sigma_(i=1)^(n) (x_(i)-barx)

If the mean of n observations x_1,x_2,x_3...x_n is barx then the sum of deviations of observations from mean is

If barx is the mean of n observations x_(1),x_(2),x_(3)……x_(n) , then the value of sum_(i=1)^(n)(x_(i)-barx) is (i) -1 (ii) 0 (iii) 1 (iv) n-1

If b_i=1-a_i ,n a=sum_(i=1)^n a_i ,n b=sum_(i=1)^n b_i ,t h e nsum_(i=1)^n a_i ,b_i+sum_(i=1)^n(a_i-a)^2= a b b. n a b c. (n+1)a b d. n a b

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