Given that $X$ has an exponential distribution with parameter $\lambda = 1$, we can solve each part using properties of the exponential distribution.

The probability density function (PDF) of the exponential distribution is given by: $f(x) = \lambda e^{-\lambda x}, \quad \text{for } x \geq 0.$ Where $\lambda$ is the rate parameter (in this case, $\lambda = 1$).

a. The expected time between two successive arrivals

The expected value $E(X)$ for an exponential distribution is: $E(X) = \frac{1}{\lambda}$ Given $\lambda = 1$, the expected time is: $E(X) = \frac{1}{1} = 1$

b. The standard deviation of the time between successive arrivals

For an exponential distribution, the standard deviation $\sigma_X$ is the same as the expected value: $\sigma_X = \frac{1}{\lambda}$ Thus, the standard deviation is: $\sigma_X = \frac{1}{1} = 1$

c. $P(X \leq 4)$

The cumulative distribution function (CDF) of the exponential distribution is: $F(x) = 1 - e^{-\lambda x}, \quad \text{for } x \geq 0.$ To find $P(X \leq 4)$, we compute the CDF at $x = 4$: $P(X \leq 4) = F(4) = 1 - e^{-1 \times 4} = 1 - e^{-4}.$ Using $e^{-4} \approx 0.0183$, we get: $P(X \leq 4) \approx 1 - 0.0183 = 0.9817.$

d. $P(2 \leq X \leq 5)$

We can find this probability using the CDF: $P(2 \leq X \leq 5) = F(5) - F(2)$ First, compute $F(5)$ and $F(2)$: $F(5) = 1 - e^{-1 \times 5} = 1 - e^{-5} \approx 1 - 0.0067 = 0.9933,$ $F(2) = 1 - e^{-1 \times 2} = 1 - e^{-2} \approx 1 - 0.1353 = 0.8647.$ Now, subtract: $P(2 \leq X \leq 5) = 0.9933 - 0.8647 = 0.1286.$

Final Answers:

• a. $E(X) = 1$
• b. $\sigma_X = 1$
• c. $P(X \leq 4) \approx 0.9817$
• d. $P(2 \leq X \leq 5) \approx 0.1286$

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Would you like more details or have any questions about these calculations?

Here are 5 related questions to expand your understanding:

1. How does the rate parameter $\lambda$ affect the shape of the exponential distribution?
2. What is the memoryless property of the exponential distribution, and how is it useful?
3. How do we compute probabilities for intervals when $\lambda$ is not 1?
4. How would the expected value change if $\lambda = 0.5$?
5. Can you explain how the CDF of an exponential distribution is derived?

Tip: The exponential distribution is widely used to model waiting times between independent events that happen at a constant average rate, like arrivals or failures.