Question

A data consists of 20 observations $x_1, x_2, \dots, x_{20}$. If $\sum_{i=1}^{20} (x_i + 5)^2 = 2500$ and $\sum_{i=1}^{20} (x_i - 5)^2 = 100$, then the ratio of mean to standard deviation of this data is:

• A. 2:1
• B. 3:1
• C. 3:2
• D. 4:1

Hint:

Expand the squared terms and subtract/add the two given summations to find the sum of observations and the sum of their squares.

Answer 1

The Correct Option is B

This problem utilizes properties of mean and variance. Let's denote the sum $\sum (x_i+5)^2$ as $S_1$ and $\sum (x_i-5)^2$ as $S_2$.
From $S_1 - S_2 = 2400$, we apply the difference of squares identity $(x_i+5)^2 - (x_i-5)^2 = 2(x_i \cdot 5 + 5 \cdot x_i) = 20x_i$.
Thus, $\sum 20x_i = 2400$, which leads to $\sum x_i = 120$.
The mean is $\mu = \frac{120}{20} = 6$.
To find the variance, consider $S_1 + S_2 = 2600$. This gives:
$$ \sum [(x_i+5)^2 + (x_i-5)^2] = 2600 $$
$$ \sum (2x_i^2 + 50) = 2600 \implies 2 \sum x_i^2 + 1000 = 2600 \implies \sum x_i^2 = 800 $$
The variance is $\sigma^2 = \frac{1}{n}\sum x_i^2 - \mu^2$. Substituting values:
$$ \sigma^2 = \frac{800}{20} - 6^2 = 40 - 36 = 4 $$
Therefore, the standard deviation $\sigma = \sqrt{4} = 2$.
The required ratio is $\frac{\mu}{\sigma} = \frac{6}{2} = 3$. This corresponds to 3:1.