Given that $X_1, X_2, \dots, X_{10}$ are independent and identically distributed (iid) random variables following an exponential distribution with rate parameter $\lambda$, and the sample mean $\bar{X} = 2.0$, we are tasked with finding the maximum likelihood estimate (MLE) of $P(X \leq 3)$.

Step 1: Maximum Likelihood Estimate of $\lambda$

For an exponential distribution $X \sim \text{Exp}(\lambda)$, the probability density function (PDF) is:

$f_X(x) = \lambda e^{-\lambda x}, \quad x \geq 0$

The mean of an exponential distribution is given by:

$\mathbb{E}(X) = \frac{1}{\lambda}$

Given that the sample mean $\bar{X} = 2.0$, the MLE for $\lambda$ is obtained by setting the sample mean equal to the theoretical mean:

$\bar{X} = \frac{1}{\lambda}$

Thus, solving for $\lambda$:

$\lambda = \frac{1}{\bar{X}} = \frac{1}{2.0} = 0.5$

Step 2: Computing $P(X \leq 3)$

Now, we want to calculate the probability $P(X \leq 3)$ under the MLE of $\lambda$. For an exponential distribution, the cumulative distribution function (CDF) is:

$F_X(x) = P(X \leq x) = 1 - e^{-\lambda x}$

Substituting $\lambda = 0.5$ and $x = 3$:

$P(X \leq 3) = 1 - e^{-0.5 \times 3} = 1 - e^{-1.5}$

Now, evaluating the expression:

$P(X \leq 3) \approx 1 - 0.2231 = 0.7769$

Conclusion

The maximum likelihood estimate of $P(X \leq 3)$ is approximately 0.7769.

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Would you like more details on the exponential distribution or MLE? Feel free to ask!

Here are 5 related questions to explore further:

1. How does the exponential distribution model real-life phenomena?
2. What are the key properties of the exponential distribution?
3. How do you derive the likelihood function for exponential distributions?
4. What is the significance of the rate parameter $\lambda$ in an exponential distribution?
5. How do you compute confidence intervals for the parameter $\lambda$?

Tip: When dealing with exponential distributions, the mean is a simple and intuitive estimator for the rate parameter $\lambda$, making the MLE calculation straightforward.