Statistics
Standard Deviation
Grade 11
Question:
<p>A data consists of \(n\) observations: \(x_1, x_2, ..., x_n\). If \(\displaystyle\sum_{i=1}^{n}(x_i + 1)^2 = 9n\) and \(\displaystyle\sum_{i=1}^{n}(x_i - 1)^2 = 5n\), then the standard deviation of this data is:</p>
<p>2</p>
<p>\(\sqrt{5}\)</p>
<p>5</p>
<p>\(\sqrt{7}\)</p>
Step-by-Step Solution
Key Concept: Expand both sum of squares equations and subtract them to find the mean, then use either equation to find variance. The relationship between the two equations directly reveals mean and variance without computing individual deviations.
<p><strong>Step 1:</strong> Expand both equations using the formula for sum of squares.</p><p>$$\sum_{i=1}^{n}(x_i + 1)^2 = \sum_{i=1}^{n}(x_i^2 + 2x_i + 1) = \sum x_i^2 + 2\sum x_i + n = 9n$$</p><p>$$\sum_{i=1}^{n}(x_i - 1)^2 = \sum_{i=1}^{n}(x_i^2 - 2x_i + 1) = \sum x_i^2 - 2\sum x_i + n = 5n$$</p><p><strong>Step 2:</strong> Subtract the second equation from the first.</p><p>$$\left(\sum x_i^2 + 2\sum x_i + n\right) - \left(\sum x_i^2 - 2\sum x_i + n\right) = 9n - 5n$$</p><p>$$4\sum x_i = 4n$$</p><p>$$\sum x_i = n \implies \bar{x} = 1$$</p><p><strong>Step 3:</strong> Substitute $\bar{x} = 1$ back into the first equation to find $\sum x_i^2$.</p><p>$$\sum x_i^2 + 2n + n = 9n$$</p><p>$$\sum x_i^2 = 6n$$</p><p><strong>Step 4:</strong> Calculate variance using $\text{Var} = \frac{\sum x_i^2}{n} - \bar{x}^2$.</p><p>$$\text{Var} = \frac{6n}{n} - 1^2 = 6 - 1 = 5$$</p><p><strong>Step 5:</strong> Standard deviation is the square root of variance.</p><p>$$\sigma = \sqrt{5}$$</p><p>∴ Answer: D</p>
Correct Answer: D