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Miscellaneous Exercise 8 (II) · Q44

Q.The arithmetic mean and standard deviation of a series of 20 items were calculated by a student as 20 cms and 5 cms respectively. But while calculating them, an item 13 was misread as 30. Find the corrected mean and standard deviation.

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Step 1: Original: n=20n=20, mean =20=20, so original (wrong) total =20×20=400=20\times20=400; this total wrongly includes 30 where it should include the true value 13. Corrected total =400−30+13=383=400-30+13=383, so corrected mean =383/20=19.15=383/20=19.15.

Step 2: Original S.D. =5=5, so original (wrong) Var=25\text{Var}=25, giving original (wrong) ∑x2=n(Var+mean2)=20(25+400)=8500\sum x^2 = n(\text{Var}+\text{mean}^2)=20(25+400)=8500; this wrongly includes 302=90030^2=900 where it should include 132=16913^2=169. Corrected ∑x2=8500−900+169=7769\sum x^2 = 8500-900+169=7769. …

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