If for distribution of 18 observations `sum(x_i-5)=3a n dsum(x_i-5)^2=43 ,` find the mean and standard deviation.

If, s is the standard deviation of the observations x_(1),x_(2),x_(3),x_(4) and x_(5) then the standard deviation of the observations kx_(1),kx_(2),kx_(3),kx_(4) and kx_(5) is

If for a distribution sum(x-5)=3,sum(x-5)^(2)=43 and the total number of terms is 18 then mean and variance are

A sample of 35 observations has mean 80 men and standard deviation 4. A second sample of 65 observations has mean 70 and standard deviation =3. Find the combined mean and standard deviation.

The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

The standard deviation of a distribution is 30. If each observation is increased by 5, then the new standard deviation will be

Consider the following statements : (i) Mean of 100 observations is 50 and standard deviation is 10. If 5 is added to each observation the new mean and standard deviation are 55, 10. (ii) Mean of 100 observations is 50 and standard deviation is 10. If each observation is multiplied by 3 then the new mean and standard deviation are 50, 10/3. The true statements are :

If the sum of the squares of deviations for 10 observations taken from their means is 2.5, then write the value of standard deviation.

If mean of squares of deviations of a set of n observations about -2a n d2 are 18 and 10 respectively then standard deviation of this set of observations is 3 (b) 2 (c) 1 (d) None of these

If mean of squares of deviations of a set of n observations about -2a n d2 are 18 and 10 respectively then standard deviation of this set of observations is 3 (b) 2 (c) 1 (d) None of these

A standard obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. Later it was discovered that he had wrongly copied down and observation as 50 instead of 40. Calculate the true mean and standard deviation.