Let the value of a random sample drawn from a normal distribution with mean 5 and unknown standard deviation $\sigma$ be 4.8, 4.5, 5.1, 5.2, 5.3, 5.5. Then, the maximum likelihood estimate of $\sigma^2$ is _____. (rounded off to two decimal places).
Correct Answer: 0.1
Solution and Explanation
Maximum Likelihood Estimate (MLE) of Variance
Step 1: Formula for Maximum Likelihood Estimate of Variance
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:
- \( x_i \) are the sample values
- \( \bar{x} \) is the sample mean
- \( n \) is the sample size
Step 2: Calculating the Sample Mean
The given sample values are:
\[ 4.8, 4.5, 5.1, 5.2, 5.3, 5.5 \]
Compute the sample mean:
\[ \bar{x} = \frac{4.8 + 4.5 + 5.1 + 5.2 + 5.3 + 5.5}{6} \]
\[ = \frac{30.4}{6} = 5.0667 \]
Step 3: Calculating the Squared Deviations
Compute the squared deviations from the mean:
- \( (4.8 - 5.0667)^2 = 0.0711 \)
- \( (4.5 - 5.0667)^2 = 0.3181 \)
- \( (5.1 - 5.0667)^2 = 0.0011 \)
- \( (5.2 - 5.0667)^2 = 0.0001 \)
- \( (5.3 - 5.0667)^2 = 0.0001 \)
- \( (5.5 - 5.0667)^2 = 0.1876 \)
Step 4: Compute the MLE of Variance
\[ \hat{\sigma}^2 = \frac{0.0711 + 0.3181 + 0.0011 + 0.0001 + 0.0001 + 0.1876}{6} \]
\[ = \frac{0.5781}{6} = 0.10 \]
Final Answer:
The maximum likelihood estimate of variance is 0.10.
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