Statistics And Probability Question 546
Question: The S.D. of a variate x is s. The S.D. of the variate $ \frac{ax+b}{c} $ where a, b, c are constant, is
[Pb. CET 1996]
Options:
A) $ ( \frac{a}{c} ),\sigma $
B) $ | \frac{a}{c} |,\sigma $
C) $ ( \frac{a^{2}}{c^{2}} ),\sigma $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ y=\frac{ax+b}{c} $ i.e., $ y=\frac{a}{c}x+\frac{b}{c} $ i.e., $ y=Ax+B $ , where $ A=\frac{a}{c} $ , $ B=\frac{b}{c} $ \ $ \bar{y}=A\bar{x}+B $ \ $ y-\bar{y}=A(x-\bar{x}) $
Þ $ {{(y-\bar{y})}^{2}}=A^{2}{{(x-\bar{x})}^{2}} $
Þ $ \sum {{(y-\bar{y})}^{2}}=A^{2}\sum {{(x-\bar{x})}^{2}} $
Þ $ n.\sigma _y^{2}=A^{2}.n\sigma _x^{2} $
Þ $ \sigma _y^{2}=A^{2}\sigma _x^{2} $
Þ $ {\sigma _{y}}=|A|{\sigma _{x}} $
Þ $ {\sigma _{y}}=| \frac{a}{c} |{\sigma _{x}} $ Thus, new S.D. $ =| \frac{a}{c} |,\sigma $ .
BETA
