Question 1167858: Many electronics follow a failure rate described by an exponential probability density function (PDF). Solar panels are advertised to last 20 years or longer, but panels made in China are failing at a higher rate. The time-to-failure of this device is usually exponentially distributed with mean 10 years.
What is the probability of failure in the first 9 years?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! The time-to-failure of the device is exponentially distributed with a mean of 10 years.
For an exponential distribution:
* The mean ($\mu$) is given by $1/\lambda$.
* The cumulative distribution function (CDF), which gives the probability of failure by time $t$, is $P(T \le t) = 1 - e^{-\lambda t}$.
Given the mean $\mu = 10$ years:
$\mu = 1/\lambda \implies 10 = 1/\lambda \implies \lambda = 1/10 = 0.1$.
We want to find the probability of failure in the first 9 years, which means $P(T \le 9)$.
Using the CDF formula with $t=9$ and $\lambda=0.1$:
$P(T \le 9) = 1 - e^{-(0.1)(9)}$
$P(T \le 9) = 1 - e^{-0.9}$
Now, we calculate the value:
$e^{-0.9} \approx 0.40657$
$P(T \le 9) = 1 - 0.40657$
$P(T \le 9) = 0.59343$
The probability of failure in the first 9 years is approximately **0.5934**.
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