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Question
The mean of 100 items was found to be 30. If at the time of calculation two items were wrongly taken as 32 and 12 instead of 23 and 11, find the correct mean.
Solution
Here,
n = 100, X = 30.
So, `bar"X" = (1)/"n"(sumx_i)`
⇒ `sumx_i = "n"bar"X"`
⇒ `sumx_i` = 100 x 30 = 3000.
∴ Incorrect value of" `sumx_i` = 3000
Now, correct value of
`sumx_i = "Incorrect value of" sumx_i` - (Sum of incorrect values) + (Sum of correct values)
= 3000 - (32 + 12) + (23 + 11)
= 2290.
∴ Correct mean
= `("Correct value of" sumx_i)/"n"`
= `(2990)/(100)`
= 29.9.
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