Question

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).

Answer 1

Correct Answer: 0.1

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.