The mean and standard deviation of 100 observ | vedclass

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(D) Given: $n = 100$,$\bar{x} = 40$,$\sigma = 5.1$.We know that $	ext{Variance} = \sigma^2 = (5.1)^2 = 26.01$.The formula for variance is $\sigma^2 =

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$161701$

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(D) Given: $n = 100$,$\bar{x} = 40$,$\sigma = 5.1$.We know that $	ext{Variance} = \sigma^2 = (5.1)^2 = 26.01$.The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.Substituting the values: $26.01 = \frac{\sum x_i^2}{100} - (40)^2$.$26.01 = \frac{\sum x_i^2}{100} - 1600$.$\frac{\sum x_i^2}{100} = 1626.01$.$\sum x_i^2 = 162601$ (This is the incorrect sum of squares).To find the correct sum of squares,we subtract the square of the incorrect observation and add the square of the correct observation:$	ext{Correct } \sum x_i^2 = 162601 - (50)^2 + (40)^2$.$	ext{Correct } \sum x_i^2 = 162601 - 2500 + 1600$.$	ext{Correct } \sum x_i^2 = 161701$.

The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$ respectively. Later on,it was found that one of the observations was taken as $50$ in place of $40$. If the wrong entry is replaced by the correct one,then the sum of the squares of all the observations is:

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