Question

The mean of the $ 100 $ observations is $ 50 $ and their standard deviation is $ 5 $ . The sum of the squares of all the observations is:
A. $ 50000 $
B. $ 250000 $
C. $ 252500 $
D. $ 255000 $

Answer 1

Hint: The sum of the squares of the observations is equal to the sum of the variance multiplied with the number of terms and the mean of the observations times the number of observations given. The variance of the distribution is equal to the square of the standard deviation of the data.

Complete step-by-step answer:
The mean of the $ 100 $ observations is $ 50 $ and their standard deviation is $ 5 $ .
As per given in the question the standard deviation of $ 100 $ observations is $ 5 $ . The variance of the data is always equal to the square of the standard deviation of the data.
The variance of the data is equal to $ {\sigma ^2} = {5^2} = 25 $ .
Now according to the formula the sum of the squares of the observations is equal to the sum of the variance multiplied with number of terms and the mean of the observations times the number of observations given i.e.
$ {\sigma ^2} = \dfrac{1}{n}\left( ({\sum {x_i^2) - {{n\times \overline X }^2}} } \right) $ .
As per given in the question, the mean of the data is equal to $ 50 $ i.e. $ \overline X = 50 $ and the number of terms is equal to $ n = 100 $ .
Substitute the value of the variance, mean and number of observations in the formula:
 $
  {\sigma ^2} = \dfrac{1}{n}\left(( {\sum\limits_{i = 1}^{100} {x_i^2) - {{N\times \overline X }^2}} } \right) \\
  25 = \dfrac{1}{{100}}\left( {\sum\limits_{i = 1}^{100} {x_i^2} - 100{{\left( {50} \right)}^2}} \right) \\
  2500 = \sum\limits_{i = 1}^{100} {x_i^2} - 250000 \\
  \sum\limits_{i = 1}^{100} {x_i^2} = 250000 + 2500 \\
  \sum\limits_{i = 1}^{100} {x_i^2} = 252500 \;
  $
So, the sum of the squares of the observations is equal to $ 252500 $ .
So, the correct answer is “Option C”.

Note: The variance of the data is always equal to the square of the standard deviation of the given data and the formula used for finding the sum of the squares of terms is given by $ {\sigma ^2} = \dfrac{1}{n}\left( {\sum {x_i^2 - {{\overline X }^2}} } \right) $ .