document.write( "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.
\n" ); document.write( "What is the probability of failure in the first 9 years? \n" ); document.write( "

Algebra.Com's Answer #852086 by CPhill(1959) $\"\"$ $\"About$

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The time-to-failure of the device is exponentially distributed with a mean of 10 years.\r
\n" ); document.write( "\n" ); document.write( "For an exponential distribution:
\n" ); document.write( "* The mean ($\mu$) is given by $1/\lambda$.
\n" ); document.write( "* The cumulative distribution function (CDF), which gives the probability of failure by time $t$, is $P(T \le t) = 1 - e^{-\lambda t}$.\r
\n" ); document.write( "\n" ); document.write( "Given the mean $\mu = 10$ years:
\n" ); document.write( "$\mu = 1/\lambda \implies 10 = 1/\lambda \implies \lambda = 1/10 = 0.1$.\r
\n" ); document.write( "\n" ); document.write( "We want to find the probability of failure in the first 9 years, which means $P(T \le 9)$.
\n" ); document.write( "Using the CDF formula with $t=9$ and $\lambda=0.1$:
\n" ); document.write( "$P(T \le 9) = 1 - e^{-(0.1)(9)}$
\n" ); document.write( "$P(T \le 9) = 1 - e^{-0.9}$\r
\n" ); document.write( "\n" ); document.write( "Now, we calculate the value:
\n" ); document.write( "$e^{-0.9} \approx 0.40657$
\n" ); document.write( "$P(T \le 9) = 1 - 0.40657$
\n" ); document.write( "$P(T \le 9) = 0.59343$\r
\n" ); document.write( "\n" ); document.write( "The probability of failure in the first 9 years is approximately **0.5934**. \n" ); document.write( "

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