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.
