Question

Let the observations \(x_i\ (1\le i\le10)\) satisfy the equations,
\(\sum_{i=1}^{10}(x_1-5)=10\) and \(\sum_{i=1}^{10}(x_1-5)^2=40\).
If \(\mu\) and \(\lambda\) are the mean and the variance of the observations, \(x_1-3,x_2-3,\dots,x_{10}-3\), then the ordered pair \((\mu,\lambda)\) is equal to :

• A. \((6,6)\)
• B. \((3,3)\)
• C. \((3,6)\)
• D. \((6,3)\)

Hint:

Let y_i = x_i - 5; then x_i - 3 = y_i + 2, and variance is unaffected by a shift.

Answer 1

The Correct Option is B

Treat the sums as over \(x_i\) (the printed subscript 1 is a typo for i). Let \(y_i=x_i-5\): \(\sum y_i=10\), \(\sum y_i^2=40\).
The new observations are \(x_i-3=y_i+2\).
Mean \(\mu=\frac{\sum y_i}{10}+2=1+2=3\).
Variance \(\lambda=\mathrm{Var}(y_i)=\frac{40}{10}-\left(\frac{10}{10}\right)^2=4-1=3\).
\((\mu,\lambda)=(3,3)\).
Answer: (2) (3, 3)