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Question
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
Options
50,000
250,000
252500
255000
Solution
Let \[\bar{ x} \] and \[\sigma\] be the mean and standard deviation of 100 observations, respectively.
\[\therefore x = 50, \sigma = 5\] and n = 100
Mean,\[\bar{ x} \] = 50
\[\Rightarrow \frac{\sum_{} x_i}{100} = 50\]
\[ \Rightarrow \sum_{} x_i = 5000 . . . . . \left( 1 \right)\]
Now,
Standard deviation,
\[\sigma = 5\]
\[\Rightarrow \sqrt{\frac{\sum_{} x_i^2}{100} - \left( \frac{\sum_{} x_i}{100} \right)^2} = 5\]
\[ \Rightarrow \frac{\sum_{} x_i^2}{100} - \left( \frac{5000}{100} \right)^2 = 25 \left[ \text{ From } \left( 1 \right) \right]\]
\[ \Rightarrow \frac{\sum_{} x_i^2}{100} = 25 + 2500 = 2525\]
\[ \Rightarrow \sum_{} x_i^2 = 252500\]
Thus, the sum of all squares of all the observations is 252500.
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