Question:

The lifetime of a component is exponentially distributed with parameter 2. The probability that its lifetime exceeds the expected lifetime (rounded to 2 decimal places) is \(\underline{\hspace{2cm}}\).

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For an exponential distribution, the probability of exceeding the mean is always \( e^{-1} \), independent of the parameter.
Updated On: Jan 30, 2026
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Correct Answer: 0.35

Solution and Explanation

Step 1: Recall properties of exponential distribution.
For an exponential distribution with parameter \( \lambda = 2 \), the expected value is:
\[ E(X) = \frac{1}{\lambda} = \frac{1}{2} \]

Step 2: Use the survival function.
The probability that the lifetime exceeds a value \( t \) is:
\[ P(X > t) = e^{-\lambda t} \]

Step 3: Substitute the expected lifetime.
\[ P\left(X > \frac{1}{2}\right) = e^{-2 \times \frac{1}{2}} = e^{-1} \]

Step 4: Numerical approximation.
\[ e^{-1} \approx 0.3679 \]

Step 5: Round to two decimal places.
\[ P(X > E(X)) \approx 0.39 \] % Final Answer

Final Answer: \[ \boxed{0.39} \]

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