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IRINA_888 [86]
2 years ago
9

Let x denote the lifetime of a mcchine component with an exponential distribution. The mean time for the component failure is 25

00 hours. What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?
Mathematics
1 answer:
aliina [53]2 years ago
4 0

Answer:

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The mean time for the component failure is 2500 hours.

This means that m = \frac{2500}, \mu = \frac{1}{2500} = 0.0004

What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

The standard deviation of the exponential distribution is the same as the mean, so this is P(X > 5000).

P(X > x) = e^{-0.0004*5000} = 0.1353

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

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