<Dl> <Dd> 1 N ∑ i = 1 N (x i − x _̄) 2 = 1 N (∑ i = 1 N x i 2) − x _̄ 2 = (1 N ∑ i = 1 N x i 2) − (1 N ∑ i = 1 N x i) 2 . (\ displaystyle (\ sqrt ((\ frac (1) (N)) \ sum _ (i = 1) ^ (N) (x_ (i) - (\ overline (x))) ^ (2))) = (\ sqrt ((\ frac (1) (N)) \ left (\ sum _ (i = 1) ^ (N) x_ (i) ^ (2) \ right) - (\ overline (x)) ^ (2))) = (\ sqrt (\ left ((\ frac (1) (N)) \ sum _ (i = 1) ^ (N) x_ (i) ^ (2) \ right) - \ left ((\ frac (1) (N)) \ sum _ (i = 1) ^ (N) x_ (i) \ right) ^ (2))).) </Dd> </Dl> <Dd> 1 N ∑ i = 1 N (x i − x _̄) 2 = 1 N (∑ i = 1 N x i 2) − x _̄ 2 = (1 N ∑ i = 1 N x i 2) − (1 N ∑ i = 1 N x i) 2 . (\ displaystyle (\ sqrt ((\ frac (1) (N)) \ sum _ (i = 1) ^ (N) (x_ (i) - (\ overline (x))) ^ (2))) = (\ sqrt ((\ frac (1) (N)) \ left (\ sum _ (i = 1) ^ (N) x_ (i) ^ (2) \ right) - (\ overline (x)) ^ (2))) = (\ sqrt (\ left ((\ frac (1) (N)) \ sum _ (i = 1) ^ (N) x_ (i) ^ (2) \ right) - \ left ((\ frac (1) (N)) \ sum _ (i = 1) ^ (N) x_ (i) \ right) ^ (2))).) </Dd> <P> This means that the standard deviation is equal to the square root of the difference between the average of the squares of the values and the square of the average value . See computational formula for the variance for proof, and for an analogous result for the sample standard deviation . </P> <P> A large standard deviation indicates that the data points can spread far from the mean and a small standard deviation indicates that they are clustered closely around the mean . </P>

How do you find the sd in statistics