Question

Let \(x_1, x_2, \ldots, x_{11}\) be the observations satisfying \(\sum_{i=1}^{11} (x_i - 4) = 22\) and \(\sum_{i=1}^{11} (x_i - 4)^2 = 154\). If the mean and variance of the observations are \(\alpha\) and \(\beta\), then the quadratic equation having the roots \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) is:

• A. \(15x^2 - 16x + 15 = 0 \)
• B. \(15x^2 - 34x + 15 = 0 \)
• C. \(x^2 - 16x + 60 = 0 \)
• D. \(12x^2 - 25x + 20 = 0 \)

Hint:

For grouped data sums like \(\sum (x_i - A)\) and \(\sum (x_i - A)^2\), it's often helpful to use a change of variable \(y_i = x_i - A\). Remember that adding/subtracting a constant to observations changes the mean by that constant, but does not affect the variance. The formula for variance is \(Var(X) = E(X^2) - (E(X))^2\), or for a discrete distribution \(\frac{\sum X^2}{N} - \left(\frac{\sum X}{N}\right)^2\).

Answer 1

The Correct Option is B

Step 1: Simplify the given sums by a change of variable.
Let \(y_i = x_i - 4\).
We are given:
Number of observations, \(N = 11\).
Sum of the new observations: \(\sum_{i=1}^{11} (x_i - 4) = \sum_{i=1}^{11} y_i = 22\).
Sum of squares of the new observations: \(\sum_{i=1}^{11} (x_i - 4)^2 = \sum_{i=1}^{11} y_i^2 = 154\).
Step 2: Calculate the mean and variance of \(y_i\). The mean of \(y_i\) is \(\bar{y}\): \[ \bar{y} = \frac{\sum y_i}{N} = \frac{22}{11} = 2 \] The variance of \(y_i\) is \(Var(y)\): \[ Var(y) = \frac{\sum y_i^2}{N} - (\bar{y})^2 \] \[ Var(y) = \frac{154}{11} - (2)^2 \] \[ Var(y) = 14 - 4 = 10 \] Step 3: Relate the mean and variance of \(y_i\) to those of \(x_i\).
If \(y_i = x_i - C\), then:
The mean of \(x_i\) is \(\bar{x} = \bar{y} + C\).
The variance of \(x_i\) is \(Var(x) = Var(y)\).
In this case, \(C = 4\).
The mean of the observations \(x_i\) is \(\alpha\): \[ \alpha = \bar{x} = \bar{y} + 4 = 2 + 4 = 6 \] The variance of the observations \(x_i\) is \(\beta\): \[ \beta = Var(x) = Var(y) = 10 \] Step 4: Determine the roots of the quadratic equation.
The roots of the quadratic equation are given as \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\).
The roots are \(\frac{6}{10}\) and \(\frac{10}{6}\).
Simplify the roots: \(\frac{3}{5}\) and \(\frac{5}{3}\).
Step 5: Form the quadratic equation.
A quadratic equation with roots \(r_1\) and \(r_2\) is given by \(x^2 - (r_1 + r_2)x + (r_1 r_2) = 0\). Calculate the sum of the roots: \[ S = \frac{3}{5} + \frac{5}{3} = \frac{3 \times 3 + 5 \times 5}{5 \times 3} = \frac{9 + 25}{15} = \frac{34}{15} \] Calculate the product of the roots: \[ P = \left(\frac{3}{5}\right) \times \left(\frac{5}{3}\right) = 1 \] Now, substitute these values into the quadratic equation formula: \[ x^2 - \left(\frac{34}{15}\right)x + 1 = 0 \] To remove the fraction, multiply the entire equation by 15: \[ 15x^2 - 34x + 15 = 0 \]