Consider a finite sequence of random values \(X=[x_1,x_2,\dots x_n]\). Let…
2011
Consider a finite sequence of random values \(X=[x_1,x_2,\dots x_n]\). Let \(\mu_x\) be the mean and \(\sigma_x\) be the standard deviation of \(X\). Let another finite sequence \(Y\) of equal length be derived from this as \(y_i=a*x_i+b\), where \(a\) and \(b\) are positive constants. Let \(\mu_y\) be the mean and \(\sigma_y\) be the standard deviation of this sequence.
Which one of the following statements is INCORRECT?
- A.
Index position of mode of
\(X\)in\(X\)is the same as the index position of mode of\(Y\)in\(Y\) - B.
Index position of median of
\(X\)in\(X\)is the same as the index position of median of\(Y\)in\(Y\) - C.
\(\mu_y=a \mu_x + b\) - D.
\(\sigma_y=a \sigma_x + b\)
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Correct answer: D
We have y_i = a x_i + b with a, b > 0. Key properties:
Mean: mu_y = a mu_x + b. Derivation: mu_y = (1/n) sum y_i = (1/n) sum (a x_i + b) = a mu_x + b.
Standard deviation: sigma_y = |a| sigma_x. Derivation: sigma_y = sqrt((1/n) sum (y_i - mu_y)^2) = sqrt((1/n) sum (a (x_i - mu_x))^2) = |a| sigma_x. With a > 0 this simplifies to sigma_y = a sigma_x.
Mode and median indices: A positive linear transformation is monotonic and one-to-one, so it preserves order and multiplicities. Therefore the positions in the sequence of the mode(s) and the median element remain the same after transforming each element by y_i = a x_i + b.
Conclusion: The statement asserting that the standard deviation transforms as sigma_y = a sigma_x + b is incorrect. The correct relation is sigma_y = a sigma_x (since a > 0), and the +b term should not appear.