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

44.0

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44.46

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45.00

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45.54

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Solution

The correct option is D $45.54$
$45 = \frac{X_{1} + X_{2} + . . . . . + X_{98} + 19 + 31}{100} \Rightarrow 98 \sum i = 1 X_{i} = 4450$
$∴$ Required Mean $= \frac{\sum_{i = 1}^{98} X_{i} + 91 + 13}{100} = \frac{4450 + 91 + 13}{100}$
$= 45.54$

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

The correct option is D 45.5445=X1+X2+.....+X98+19+31100⇒98∑i=1Xi=4450∴ Required Mean =∑98i=1Xi+91+13100=4450+91+13100=45.54

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

The correct option is D $45.54$
$45 = \frac{X_{1} + X_{2} + . . . . . + X_{98} + 19 + 31}{100} \Rightarrow 98 \sum i = 1 X_{i} = 4450$
$∴$ Required Mean $= \frac{\sum_{i = 1}^{98} X_{i} + 91 + 13}{100} = \frac{4450 + 91 + 13}{100}$
$= 45.54$