A set of n values `x_(1),x_(2),……,x_(n)` has standard deviation `sigma`. What is the standard deviation of n values `x_(1)+k, x_(2)+k….,x_(n)+k?`

A set of n values x_(1),x_(2),…….,x_(n) has standard deviation sigma . The standard deviation of n values x_(1)-k,x_(2)-k,…….,x_(n)-k is

A set of n values x_(1), x_(2), ….., x_(n) has standard deviation sigma . The standard deviation of n values : x_(1) + k, x_(2) + k, ……, x_(n) + k will be :

A set of n value x_(1),x_(2),……..,x_(n) has mean barx and standard deviation sigma . The mean and standard deviation of n values (x_(1))/(k),(x_(2))/(k),.......,(x_(n))/(k)(kne0 respectively are (i) kbarx,(sigma)/(k) (ii) (barx)/(k),(sigma)/(k) (iii) kbarx,ksigma (iv) (barx)/(k),ksigma

The standard deviation of n observation x_(1),x_(2),……x_(n) is 6. The standard deviation of another set of n observations y_(1),y_(2),………..y_(n) is 8. What is the standard deviation of n observations x_(1)-y_(2),x_(2)-y_(2),………..,x_(n)-y_(n) ?

Let x_(1),x_(2),x_(3),……..,x_(n) be n observations with mean barx and standard deviation sigma . The mean the standard deviation of kx_(1),kx_(2),……,kx_(n) respectively are (i) barx,ksigma (ii) kbarx,sigma (iii) kbarx,ksigma (iv) barx,sigma

If the standard deviation of n observation x_(1), x_(2),…….,x_(n) is 5 and for another set of n observation y_(1), y_(2),………., y_(n) is 4, then the standard deviation of n observation x_(1)-y_(1), x_(2)-y_(2),………….,x_(n)-y_(n) is

If the standard deviation of n observation x_(1), x_(2),…….,x_(n) is 5 and for another set of n observation y_(1), y_(2),………., y_(n) is 4, then the standard deviation of n observation x_(1)-y_(1), x_(2)-y_(2),………….,x_(n)-y_(n) is

If the standard deviation of x_(1),x_(2),….,x_(n) is 3.5 then the standard deviation of -2x_(1)-3, -2x_(2)-3,….,-2x_(n)-3 is