The mean and sample standard deviation of the dataset consisting of 10 observations is 16 and 10 respectively. Later it is noted that one observation 16 is wrongly noted as 10. What is the sample variance of the original dataset? (Correct to 2 decimal place accuracy)
Question
The mean and sample standard deviation of the dataset consisting of 10 observations is 16 and 10 respectively. Later it is noted that one observation 16 is wrongly noted as 10. What is the sample variance of the original dataset? (Correct to 2 decimal place accuracy)
Solution
Step 1: Calculate the sum of the dataset
The sum of the dataset can be calculated by multiplying the mean by the number of observations.
So, Sum = Mean * Number of observations = 16 * 10 = 160
Step 2: Correct the error in the dataset
The sum of the dataset after correcting the error is the original sum minus the incorrect observation plus the correct observation.
So, Corrected Sum = Original Sum - Incorrect observation + Correct observation = 160 - 10 + 16 = 166
Step 3: Calculate the corrected mean
The corrected mean is the corrected sum divided by the number of observations.
So, Corrected Mean = Corrected Sum / Number of observations = 166 / 10 = 16.6
Step 4: Calculate the sum of squares of the dataset
The sum of squares of the dataset can be calculated by adding the square of the standard deviation to the square of the mean, multiplied by the number of observations.
So, Sum of squares = (Standard deviation)^2 * Number of observations + Mean^2 * Number of observations = 10^2 * 10 + 16^2 * 10 = 1000 + 2560 = 3560
Step 5: Calculate the corrected sum of squares
The corrected sum of squares is the original sum of squares minus the square of the incorrect observation plus the square of the correct observation.
So, Corrected Sum of squares = Original Sum of squares - Incorrect observation^2 + Correct observation^2 = 3560 - 10^2 + 16^2 = 3560 - 100 + 256 = 3716
Step 6: Calculate the sample variance
The sample variance is the corrected sum of squares divided by the number of observations minus 1, minus the square of the corrected mean.
So, Sample Variance = (Corrected Sum of squares / (Number of observations - 1)) - Corrected Mean^2 = (3716 / (10 - 1)) - 16.6^2 = 412.89 - 275.56 = 137.33
So, the sample variance of the original dataset is 137.33 (correct to 2 decimal places).
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