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

44.2

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44.4

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44.42

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44.46

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Solution

The correct option is D $44.46$
$n = 100$ , $¯ ¯ ¯ x = 45$
$∴$ Sum $= n ¯ ¯ ¯ x = 4500$

Now, correct sum $= 4500 - 91 - 13 + 19 + 31 = 4446$
$∴$ Correct mean $= \frac{4446}{100} = 44.46$

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Mean of 100 observations is 45. It was later found that two observations, 19 and 31 were incorrectly recorded as 91 and 13 respectively. Then the correct mean is

The correct option is D 44.46 n=100, ¯¯¯x=45 ∴ Sum =n¯¯¯x=4500 Now, correct sum =4500−91−13+19+31=4446 ∴ Correct mean =4446100=44.46

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

The correct option is D $44.46$
$n = 100$ , $¯ ¯ ¯ x = 45$
$∴$ Sum $= n ¯ ¯ ¯ x = 4500$

Now, correct sum $= 4500 - 91 - 13 + 19 + 31 = 4446$
$∴$ Correct mean $= \frac{4446}{100} = 44.46$