The mean of `n` observations is `xdot` If `k` is added to each observation, then the new mean is ` X ` (b) ` X +k` (c) ` X -k` (d) `k X `

The mean of n observations is X dot If each observation is multiplied by k , the mean of new observation is k X (b) ( X )/k (c) X +k (d) X -k

The mean of 10 observations is 15. If one observation 15 is added, then the new mean is (a) 16 (b) 11 (c) 10 (d) 15

The mean of n observations is X . If the first item is increased by 1, second by 2 and so on, then the new mean is (a) X +n (b) X +n/2 (c) X +(n+1)/2 (d) None of these

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

The mean of 100 observations is 40 and their standard deviation respectively is 10. if 5 is added to each observation then the new mean and new standard deviation will be

There are 7 observations in the data and their mean is 11. If each observation is multiplied by 2, then the mean of new observations is (a) 11 (b) 13 (c) 22 (d) 5.5

The mean of 40 observations 20 and their standard deviation is 5. If the sum of the square of the observations k, then the value of (k)/(1000) is

If the mean of the observation x , x+3 , x+5 , x+7 and x+10 is 9 , then mean of the last three observations is

Observe the following statements : (A) : 10 is the mean of a set of 7 obervations and 5 is the mean of a set of 3 observations. The mean of a combined set is 9. (R ) : If bar(x)_(i)(i = 1,2,…,k) are the means of k - series n_(i)(I = 1,2,3,…,k) respectively, then the combined or composite mean is bar(x) = (n_(1)bar(x)_(1) + n_(2) bar(x)_(2) + ...+ n_(k)bar(x_(k)))/(n_(1)+n_(2)+...+n_(k))

Mean of a certain number of observations is x . If each observation is divided by m(m!=0) and increased by n , then the mean of new observation is (a) ( x )/m+n (b) ( x )/n+m (c) x +n/m (d) x +m/n