The mean of the 100 observations is 45. It was later found that two observations 19 and 31 are incorrectly recorded as 91 and 13. Find the correct mean.

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Solution

Mean = 45

$\frac{x}{100} = 45 \Rightarrow x = 4500$

Two number (91+13)=104 incorrect

So let sum = 4500 – 104 = 4396

Add correct number 19+31 = 50

= 4396+50 = 4446

Mean = $\frac{4446}{100} = 44.46$

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The mean of the 100 observations is 45. It was later found that two observations 19 and 31 are incorrectly recorded as 91 and 13. Find the correct mean.

Mean = 45 x100=45⇒x=4500 Two number (91+13)=104 incorrect So let sum = 4500 – 104 = 4396 Add correct number 19+31 = 50 = 4396+50 = 4446 Mean = 4446100=44.46

Mean = 45

$\frac{x}{100} = 45 \Rightarrow x = 4500$

Two number (91+13)=104 incorrect

So let sum = 4500 – 104 = 4396

Add correct number 19+31 = 50

= 4396+50 = 4446

Mean = $\frac{4446}{100} = 44.46$