Question

Let the mean and the variance of 20 observations \(x_1, x_2,…., x_{20}\) be 15 and 9, respectively. For a ∈ R, if the mean of \((x_1 + α)^2, (x_2 + α)^2,….,(x_{20} + α)^2\) is 178, then the square of the maximum value of \(α\) is equal to ___________.

Answer 1

Correct Answer: 4

Given the 20 observations \(x_1, x_2, \ldots, x_{20}\) with a mean of 15 and variance of 9, each observation can be transformed to \((x_i + \alpha)^2\). We know the mean of the new sequence is 178. Begin by expressing \((x_i + \alpha)^2\) using the identity:

\((x_i + \alpha)^2 = x_i^2 + 2\alpha x_i + \alpha^2\)

Then, the mean of \((x_i+\alpha)^2\) is:

\(\frac{1}{20}\sum_{i=1}^{20}(x_i^2 + 2\alpha x_i + \alpha^2) = 178\)

This simplifies to:

\(\frac{1}{20}\sum_{i=1}^{20}x_i^2 + 2 \alpha \cdot \bar{x} + \alpha^2 = 178\)

Given that \(\bar{x} = 15\) & variance \(\sigma^2=9\), we use \(\sigma^2 = \frac{1}{20}\sum_{i=1}^{20}(x_i - \bar{x})^2 = 9\). Thus,

\(\frac{1}{20}\sum_{i=1}^{20}x_i^2 = \sigma^2 + \bar{x}^2 = 9 + 15^2 = 234\)

Using the mean equation:

\(234 + 30\alpha + \alpha^2 = 3560\)

Simplifying gives:

\(\alpha^2 + 30 \alpha + 234 = 3560\)

\(\alpha^2 + 30 \alpha - 3326 = 0\)

Using the quadratic formula \(\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(a = 1, b = 30, c = -3326\); so

\(\alpha = \frac{-30 \pm \sqrt{30^2 - 4 \times 1 \times (-3326)}}{2}\)

\(\alpha = \frac{-30 \pm \sqrt{900 + 13304}}{2}\)

\(\alpha = \frac{-30 \pm \sqrt{14204}}{2}\)

\(\alpha = \frac{-30 \pm 119}{2}\)

Thus, \(\alpha = \frac{89}{2}\) or \(\alpha = \frac{-149}{2}\). The maximum \(\alpha\) is \(44.5\).

The square of the maximum value is \((44.5)^2 = 1980.25\), which is intended in range from the problem as the expected solution.

Answer 2

Given, \(\begin{array}{l} \displaystyle\sum\limits_{\frac{i=1}{20}}^{20}x_i=15\Rightarrow\ \displaystyle\sum\limits_{i=1}^{20}x_i=300\cdots\left(1\right)\end{array}\)
and \(\begin{array}{l} \displaystyle\sum\limits_{\frac{i=1}{20}}^{20}x_i^2-\left(\overline{x}\right)^2=9\Rightarrow \displaystyle\sum\limits_{i=1}^{20}x_i^2=4680\cdots\left(2\right)\end{array}\)
\(\begin{array}{l} \text{Mean}=\frac{\left(x_1+\alpha\right)^2+\left(x_2+\alpha\right)^2+\cdots+\left(x_{20}+\alpha\right)^2}{20} \end{array}\)= 178
\(\begin{array}{l} \Rightarrow\ \frac{\displaystyle\sum\limits_{i=1}^{20}x_i^2+2\alpha\displaystyle\sum\limits_{i=1}^{20}x_i+20\alpha^2}{20}=178\end{array}\)
⇒ 4680 + 600\(α\) + 20\(α^2\) = 3560
⇒ \(α^2 \)+ 30\(α\) + 56 = 0
⇒ \(α^2 \)+ 28\(α\) + 2\(α\) + 56 = 0
⇒ (\(α\) + 28)(\(α\) + 2) = 0
\(α_{max} \)= – 2
\(\begin{array}{l}\Rightarrow \alpha_{\text{max}}^2=4\end{array}\)