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`        The mean and standard deviation of 50 items is found tobe 40 and 5 respectively. It was later found that oneitem 50 was incorrect and the correct figure is 40.Calculate the correct standard deviation. 5`
one year ago

```							Mean = 40Standard deviation = 5Number of items =50Name the items as a1,a2,,.......a50Suppose the 50th item is wrongly indicated as 50 instead of 40Sum of all observations before correction = 50 x 40 = 2000Sum of the correct observations (a1+a2....a49) = 2000-50 = 1950Sum of 50 observations = 1950 + corrected observation = 1950 + 40 = 1990Mean = 1990 / 50 = 199/5 =39.8Sum 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) – 2500Corrected sum of squares = (a1^2+a2^2+----------------a49^2)+(40^2) = 50(1616)-2500+1600 = 79,900Mean of squares = 79,900 / 50 = 1598SD^2 =Mean of squares – Corrected Mean^2 = 1598-(39.8)^2 = 13.96SD = (13.96)^1/2
```
one year ago
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