Advertisements
Advertisements
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.
Advertisements
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.
RELATED QUESTIONS
Find the mean of the first six natural numbers.
Find the mean of all factors of 10.
Find the mean of 75 numbers, if the mean of 45 of them is 18 and the mean of the remaining ones is 13.
The mean of 200 items was 50. Later on, it was discovered that two items were misread as 92 and 8 instead of 192 and 88.
Find the correct mean.
If different values of variable x are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1;
find
(i) the mean ` barx `
(ii) the value of ` sum (x_i - barx)`
The mean of 5 numbers is 18. If one number is excluded, the mean of the remaining number becomes 16. Find the excluded number.
Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if the observations, given above, be: divided by 2.
The mean of 16 numbers is 8. If 2 is added to every number, what will be the new mean?
The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean.
If `bar"X"` is the mean of n observations x1, x2, x3,..., xn then the mean of `x_1/"a", x_2/"a", x_3/"a",...,x_"n"/"a" "is" bar"X"/"a"`, where a is an non-zero number.
i.e., if each observation is divided by a non-zero number, then the mean is also divided by it.
