We are given that the individual examined 13 defective products, correctly identified 6, and misidentified 7. The number of correctly identified defective products follows a binomial distribution:

$X \sim \text{Binomial}(n = 13, \theta)$

Additionally, we are provided with a Beta prior distribution:

$\theta \sim \text{Beta}(5.2, 2.7)$

1. Posterior Distribution

For Bayesian analysis, when using a Beta prior and a Binomial likelihood, the posterior distribution for $\theta$ also follows a Beta distribution.

The prior is:

$\theta \sim \text{Beta}(\alpha_{\text{prior}} = 5.2, \beta_{\text{prior}} = 2.7)$

The likelihood, which is binomial, adds information to update our prior distribution. After observing the data, the updated posterior distribution becomes:

$\theta \mid X \sim \text{Beta}(\alpha_{\text{post}}, \beta_{\text{post}})$

Where the posterior parameters are updated as:

$\alpha_{\text{post}} = \alpha_{\text{prior}} + \text{number of successes}$ $\beta_{\text{post}} = \beta_{\text{prior}} + \text{number of failures}$

From the data, we have:

• Number of successes = 6 (correctly identified defective products)
• Number of failures = 7 (misidentified defective products)

Thus, the posterior parameters are:

$\alpha_{\text{post}} = 5.2 + 6 = 11.2$ $\beta_{\text{post}} = 2.7 + 7 = 9.7$

Therefore, the posterior distribution is:

$\theta \mid X \sim \text{Beta}(11.2, 9.7)$

2. Posterior Expected Value

The expected value of a Beta distribution $\text{Beta}(\alpha, \beta)$ is given by:

$\mathbb{E}[\theta] = \frac{\alpha}{\alpha + \beta}$

Substituting the posterior values:

$\mathbb{E}[\theta] = \frac{11.2}{11.2 + 9.7} = \frac{11.2}{20.9} \approx 0.54$

3. Posterior Standard Deviation

The standard deviation of a Beta distribution $\text{Beta}(\alpha, \beta)$ is given by:

$\text{SD}[\theta] = \sqrt{\frac{\alpha \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)}}$

Substituting the posterior values:

$\text{SD}[\theta] = \sqrt{\frac{11.2 \times 9.7}{(11.2 + 9.7)^2 (11.2 + 9.7 + 1)}}$ $\text{SD}[\theta] = \sqrt{\frac{108.64}{20.9^2 \times 21.9}} = \sqrt{\frac{108.64}{436.81 \times 21.9}} = \sqrt{\frac{108.64}{9566.14}} \approx \sqrt{0.01136} \approx 0.1065$

Thus, the posterior standard deviation is approximately:

$\text{SD}[\theta] \approx 0.11$

Final Answer:

• Posterior expected value of $\theta$: $\approx 0.54$
• Posterior standard deviation of $\theta$: $\approx 0.11$

Would you like further details or have any other questions?

Related questions:

1. How does the prior distribution influence the posterior distribution?
2. What happens to the posterior if the prior distribution is more informative (e.g., smaller variance)?
3. How would the posterior distribution change if more defective products were tested?
4. How does the binomial likelihood combine with the beta prior mathematically?
5. How would one interpret the result of the posterior mean in a real-world context?

Tip:

In Bayesian statistics, the more data you have, the less the prior influences the posterior distribution, making the posterior rely more on the observed data.