The MLE estimate of variance for a normal distribution is given by:
\[ \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
where:
The given sample values are:
\[ 4.8, 4.5, 5.1, 5.2, 5.3, 5.5 \]
Compute the sample mean:
\[ \bar{x} = \frac{4.8 + 4.5 + 5.1 + 5.2 + 5.3 + 5.5}{6} \]
\[ = \frac{30.4}{6} = 5.0667 \]
Compute the squared deviations from the mean:
\[ \hat{\sigma}^2 = \frac{0.0711 + 0.3181 + 0.0011 + 0.0001 + 0.0001 + 0.1876}{6} \]
\[ = \frac{0.5781}{6} = 0.10 \]
The maximum likelihood estimate of variance is 0.10.
For a statistical data \( x_1, x_2, \dots, x_{10} \) of 10 values, a student obtained the mean as 5.5 and \[ \sum_{i=1}^{10} x_i^2 = 371. \] He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively.
The variance of the corrected data is: