The exponential distribution has the following key properties for a given rate parameter $\lambda$:

1. Expected Value (Mean): $E(X) = \frac{1}{\lambda}$

2. Standard Deviation: $\sigma = \frac{1}{\lambda}$

Given $\lambda = 0.2$, let's calculate these values:

1. Expected Time Between Breakdowns:

$E(X) = \frac{1}{\lambda} = \frac{1}{0.2} = 5 \, \text{days}$

2. Standard Deviation of Time Between Breakdowns:

$\sigma = \frac{1}{\lambda} = \frac{1}{0.2} = 5 \, \text{days}$

Final Answer:

• Expected Time: $5 \, \text{days}$
• Standard Deviation: $5 \, \text{days}$

Would you like more details or have any questions?

Here are 5 related questions you might explore:

1. How is the exponential distribution used in reliability analysis?
2. What is the probability that the machine breaks down within 3 days?
3. What is the cumulative distribution function (CDF) of this exponential distribution?
4. How can this information be used to plan maintenance schedules?
5. What is the memoryless property of the exponential distribution?

Tip: For exponentially distributed events, the mean and standard deviation are always the same!