mean,median

WHAT X ???

 

Did you lose it - maybe it is with the coordinate plane at the end of the bed.  

Do you think it might be CPhill ??

Find the (sample) standard deviation of the list:
(102, 121, 103, 111, 109)

The standard deviation is given by:
sqrt((variance))

The (sample) variance of a list of numbers X = {X_1, X_2, ..., X_n} with mean mu = (X_1+X_2+...+X_n)/n is given by:
(abs(X_1-mu)^2+abs(X_2-mu)^2+...+abs(X_n-mu)^2)/(n-1)

There are n = 5 elements in the list X = {102, 121, 103, 111, 109}:
(abs(X_1-mu)^2+abs(X_2-mu)^2+abs(X_3-mu)^2+abs(X_4-mu)^2+abs(X_5-mu)^2)/(5-1)

The elements X_i of the list X = {102, 121, 103, 111, 109} are:
X_1 = 102
X_2 = 121
X_3 = 103
X_4 = 111
X_5 = 109
(abs(102-mu)^2+abs(121-mu)^2+abs(103-mu)^2+abs(111-mu)^2+abs(109-mu)^2)/(5-1)

The mean (mu) is given by

mu = (X_1+X_2+X_3+X_4+X_5)/5 = (102+121+103+111+109)/5 = 546/5:
(abs(102-546/5)^2+abs(121-546/5)^2+abs(103-546/5)^2+abs(111-546/5)^2+abs(109-546/5)^2)/(5-1)

The values of the differences are:
102-546/5 = -36/5
121-546/5 = 59/5
103-546/5 = -31/5
111-546/5 = 9/5
109-546/5 = -1/5
5-1 = 4
(abs(-36/5)^2+abs(59/5)^2+abs(-31/5)^2+abs(9/5)^2+abs(-1/5)^2)/(4)

The values of the terms in the numerator are:
abs(-36/5)^2  =  1296/25
abs(59/5)^2  =  3481/25
abs(-31/5)^2  =  961/25
abs(9/5)^2  =  81/25
abs(-1/5)^2  =  1/25
(1296/25+3481/25+961/25+81/25+1/25)/4

1296/25+3481/25+961/25+81/25+1/25 = 1164/5:
291/5

The standard deviation is given by
sqrt((variance))  =  sqrt(291/5):
Answer: |  sqrt(291/5)