Yasas Chandra Sai P
Last Activity: 6 Years ago
Mean = 40
Standard deviation = 5
Number of items =50
Name the items as a1,a2,,.......a50
Suppose the 50th item is wrongly indicated as 50 instead of 40
Sum of all observations before correction = 50 x 40 = 2000
Sum of the correct observations (a1+a2....a49) = 2000-50 = 1950
Sum of 50 observations = 1950 + corrected observation = 1950 + 40 = 1990
Mean = 1990 / 50 = 199/5 =39.8
Sum of the squares of the observations = (a1^2+a2^2-------a50^2) = n(SD^2+Mean^2) = 50(4^2+40^2)
(a1^2+a2^2-----------a49^2)+(50)^2 = 50(4^2+40^2)
(a1^2+a2^2+----------a49^2) = 50(16+1600) – 2500
Corrected sum of squares = (a1^2+a2^2+----------------a49^2)+(40^2) = 50(1616)-2500+1600 = 79,900
Mean of squares = 79,900 / 50 = 1598
SD^2 =Mean of squares – Corrected Mean^2 = 1598-(39.8)^2 = 13.96
SD = (13.96)^1/2
