JEE MAIN - Mathematics (2023 - 15th April Morning Shift - No. 7)
The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is :
11
12
13
14
Explanation
1. Calculate the sum of the original observations:
$$\frac{x_1+x_2+\ldots+x_9+50}{10}=20$$
$$x_1+x_2+\ldots+x_9=150$$
2. Calculate the sum of the squares of the original observations using the original variance:
$$\sigma^2 = 8^2 = 64$$
$$64 = \frac{x_1^2+x_2^2+\ldots+x_9^2+2500}{10} - 400$$
$$x_1^2+x_2^2+\ldots+x_9^2 = 2140$$
3. Calculate the new mean after correcting the error:
$$\text{New mean} = \frac{150+40}{10} = 19$$
4. Calculate the new variance using the corrected sum of observations and the corrected sum of squares of observations:
$$\text{New } \sigma^2 = \frac{2140+1600}{10} - (19)^2$$
$$ \Rightarrow $$ $$\sigma^2 = 13$$
The correct variance after correcting the error is 13 (Option C).
$$\frac{x_1+x_2+\ldots+x_9+50}{10}=20$$
$$x_1+x_2+\ldots+x_9=150$$
2. Calculate the sum of the squares of the original observations using the original variance:
$$\sigma^2 = 8^2 = 64$$
$$64 = \frac{x_1^2+x_2^2+\ldots+x_9^2+2500}{10} - 400$$
$$x_1^2+x_2^2+\ldots+x_9^2 = 2140$$
3. Calculate the new mean after correcting the error:
$$\text{New mean} = \frac{150+40}{10} = 19$$
4. Calculate the new variance using the corrected sum of observations and the corrected sum of squares of observations:
$$\text{New } \sigma^2 = \frac{2140+1600}{10} - (19)^2$$
$$ \Rightarrow $$ $$\sigma^2 = 13$$
The correct variance after correcting the error is 13 (Option C).
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