Question

If \( \sum\limits_{i=1}^{9} (x_i - 5) = 9 \) and \( \sum\limits_{i=1}^{9} (x_i - 5)^2 = 45 \), then the standard deviation of the nine observations \( x_1, x_2, \ldots, x_9 \) is

• A. \( 2 \)
• B. \( 4 \)
• C. \( 3 \)
• D. \( 9 \)

Hint:

When you are given expressions involving deviations from a constant (like \( x_i - 5 \)), define a new variable \( y_i = x_i - 5 \). This simplifies calculations for mean and variance. Use the identity: \[ \sum (x - a)^2 = \sum x^2 - 2a\sum x + na^2 \]

Answer 1

The Correct Option is A

Step 1: Understand what's given We are given: \[ \sum_{i=1}^{9} (x_i - 5) = 9 \] \[ \sum_{i=1}^{9} (x_i - 5)^2 = 45 \] Step 2: Calculate mean Let’s define: \[ y_i = x_i - 5 \] Then: \[ \sum_{i=1}^{9} y_i = 9 \Rightarrow \text{Mean of } y_i = \frac{9}{9} = 1 \] \[ \sum_{i=1}^{9} y_i^2 = 45 \] Step 3: Use variance formula \[ \text{Variance} = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2 \Rightarrow \frac{1}{9} \sum_{i=1}^{9} (y_i - 1)^2 \] Now: \[ \sum_{i=1}^{9} (y_i - 1)^2 = \sum y_i^2 - 2 . 1 . \sum y_i + 9 . 1^2 = 45 - 18 + 9 = 36 \] So variance: \[ \frac{36}{9} = 4 \Rightarrow \text{Standard Deviation} = \sqrt{4} = 2 \]