A) \[\frac{a}{p}{{\sigma }_{x}}\]
B) \[\left| \frac{a}{p} \right|{{\sigma }_{x}}\]
C) \[\left| \frac{p}{a} \right|{{\sigma }_{x}}\]
D) \[\frac{p}{a}{{\sigma }_{x}}\]
Correct Answer: B
Solution :
[b]: Let\[y=\frac{ax+b}{P}\therefore \overline{y}=\frac{a\overline{x}+b}{P}\] Now,\[y-\overline{y}=\frac{1}{P}a(x-\overline{x})\Rightarrow {{(y-\overline{y})}^{2}}=\frac{{{a}^{2}}}{{{P}^{2}}}{{(x-\overline{x})}^{2}}\] \[\Rightarrow \] \[\frac{1}{n}\sum\limits_{{}}^{{}}{{{(y-\overline{y})}^{2}}=\frac{{{a}^{2}}}{{{p}^{2}}}\frac{1}{n}}\sum\limits_{{}}^{{}}{{{(x-\overline{x})}^{2}}}\] \[\therefore \]S.D. of \[y=\left| \frac{a}{P} \right|\]S.D. of \[x=\left| \frac{a}{P} \right|{{\sigma }_{x}}\]
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