Let X₁, X2, Xn represent a random sample from a Rayleigh distribution with the following pdf. f(x; 0) == -x²/(20) (a) Determine the maximum likelihood estimator of 0. Ox/n O (EX;) / 2n O (Ex₁²)/n Ox/2n Ο (ΣΧ. 2) / 2η Calculate the estimate from the following n = 10 observations on vibratory stress of a turbine blade under specified conditions. (Round your answer to three decimal places.) 12.38 9.01 11.00 5.04 6.31 7.56 8.68 8.25 16.38 11.86 X>0 (b) Determine the mle of the median of the vibratory stress distribution. [Hint: First express the median in terms of 0.] O 1.38630 (1.38636)³ O1.38630 (1.38636)² O O O 1.38638

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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### Educational Website Content

#### Topic: Maximum Likelihood Estimation and Vibratory Stress Distribution

---

**Problem Statement:**

Let \( X_1, X_2, \ldots, X_n \) represent a random sample from a Rayleigh distribution with the following probability density function (pdf):

\[ f(x; \theta) = \frac{x}{\theta} e^{-x^2/(2\theta)} \quad \text{ for } \quad x > 0 \]

**Tasks:**

---

**(a) Determine the maximum likelihood estimator of \( \theta \).**

Options:
- \(\circ \quad \bar{X} / n \)
- \(\circ \quad (\Sigma X_i^2) / 2n \)
- \(\circ \quad (\Sigma X_i^2) / n \)
- \(\circ \quad \bar{X} / 2n \)
- \(\circ \quad (\Sigma X_i^2) / 2n \) 

**(b) Calculate the estimate from the following \( n = 10 \) observations on vibratory stress of a turbine blade under specified conditions. (Round your answer to three decimal places.)**

Observations:
- 12.38, 9.01, 11.00, 5.04, 6.31
- 7.56, 8.68, 8.25, 16.38, 11.86

(Enter the calculated estimate in the provided box.)

**(c) Determine the maximum likelihood estimate (MLE) of the median of the vibratory stress distribution.** 
   
Hint: First, express the median in terms of \( \theta \).

Options:
- \(\circ \quad 1.3863 \hat{\theta} \)
- \(\circ \quad (1.3863 \hat{\theta})^3 \)
- \(\circ \quad 3 \sqrt{1.3863 \hat{\theta}} \)
- \(\circ \quad (1.3863 \hat{\theta})^2 \)
- \(\circ \quad 1.3863 \hat{\theta}^2 \)

---

This problem set guides you through the process of deriving the maximum likelihood estimator for the Rayleigh distribution parameter, applying the estimator to given data, and determining the MLE of the median of the distribution.
Transcribed Image Text:### Educational Website Content #### Topic: Maximum Likelihood Estimation and Vibratory Stress Distribution --- **Problem Statement:** Let \( X_1, X_2, \ldots, X_n \) represent a random sample from a Rayleigh distribution with the following probability density function (pdf): \[ f(x; \theta) = \frac{x}{\theta} e^{-x^2/(2\theta)} \quad \text{ for } \quad x > 0 \] **Tasks:** --- **(a) Determine the maximum likelihood estimator of \( \theta \).** Options: - \(\circ \quad \bar{X} / n \) - \(\circ \quad (\Sigma X_i^2) / 2n \) - \(\circ \quad (\Sigma X_i^2) / n \) - \(\circ \quad \bar{X} / 2n \) - \(\circ \quad (\Sigma X_i^2) / 2n \) **(b) Calculate the estimate from the following \( n = 10 \) observations on vibratory stress of a turbine blade under specified conditions. (Round your answer to three decimal places.)** Observations: - 12.38, 9.01, 11.00, 5.04, 6.31 - 7.56, 8.68, 8.25, 16.38, 11.86 (Enter the calculated estimate in the provided box.) **(c) Determine the maximum likelihood estimate (MLE) of the median of the vibratory stress distribution.** Hint: First, express the median in terms of \( \theta \). Options: - \(\circ \quad 1.3863 \hat{\theta} \) - \(\circ \quad (1.3863 \hat{\theta})^3 \) - \(\circ \quad 3 \sqrt{1.3863 \hat{\theta}} \) - \(\circ \quad (1.3863 \hat{\theta})^2 \) - \(\circ \quad 1.3863 \hat{\theta}^2 \) --- This problem set guides you through the process of deriving the maximum likelihood estimator for the Rayleigh distribution parameter, applying the estimator to given data, and determining the MLE of the median of the distribution.
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