A) 50.10, 51.5
B) 50.10, 52
C) 50, 51.5
D) none of these
Correct Answer: B
Solution :
[b]: Here n = 100, mean = 50 and median = 52 \[\therefore \]\[\overline{x}=\frac{1}{n}\sum\limits_{i=1}^{100}{{{x}_{i}}}=50\Rightarrow \sum\limits_{i=1}^{100}{{{x}_{i}}}=5000\] Now corrected \[\sum\limits_{i=1}^{100}{{{x}_{i}}}=5000-100+110=5010\] \[\therefore \]Corrected mean \[=\frac{1}{100}\sum\limits_{i=1}^{100}{{{x}_{i}}}=\frac{5010}{100}=51.10\] As median is positional average therefore it will remain same.
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