13.Statistics
hard

बीस प्रेक्षणों का माध्य तथा मानक विचलन क्रमश: $10$ तथा $2$ हैं। जाँच करने पर यह पाया गया कि प्रेक्षण $8$ गलत है। निम्न में से प्रत्येक का सही माध्य तथा मानक विचलन ज्ञात कीजिए यदि

उसे $12$ से बदल दिया जाए।

A

$1.98$

B

$1.98$

C

$1.98$

D

$1.98$

Solution

When $8$ is replaced by $12$

Incorrect sum of observations $=200$

$\therefore$ Correct sum of observations $=200-8+12=204$

$\therefore$ Correct mean $=\frac{\text { Correct sum }}{20}=\frac{204}{20}=10.2$

Standard deviation $\sigma  = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {{x_i}^2 – \frac{1}{{{n^2}}}{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} } $

$ = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 – {{\left( {\bar x} \right)}^2}} } $

$ \Rightarrow 2 = \sqrt {\frac{1}{{20}}Incorrect\sum\limits_{i = 1}^n {x_i^2 – {{\left( {10} \right)}^2}} } $

$ \Rightarrow 4 = \frac{1}{{20}}Incorrect\sum\limits_{i = 1}^n {x_i^2 – 100} $

$ \Rightarrow Incorrect\sum\limits_{i = 1}^n {x_i^2 = 2080} $

$\therefore Correct\,\,\sum\limits_{i = 1}^n {x_i^2 = \,} Incorrect\,\,\sum\limits_{i = 1}^n {x_i^2 – {{\left( 8 \right)}^2}} $

$=2080-64+144$

$=2160$

$\therefore$ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$

$=\sqrt{\frac{2160}{20}-(10.2)^{2}}$

$=\sqrt{108-104.04}$

$=\sqrt{3.96}$

$=1.98$

Standard 11
Mathematics

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