Question

Each time a certain machine breaks down it is replaced by a new one of the same type. In the long run, what percentage of time is the machine in use less than one year old if the life distribution of a machine is (a) uniformly distributed over \((0,2)\) ? (b) exponentially distributed with mean \(1 ?\)

Short answer

For a uniformly distributed machine, it is in use less than one year old approximately \(50\%\) of the time. For an exponentially distributed machine, it is in use less than one year old approximately \(63.21\%\) of the time.

Step-by-step solution

Part (a): Determine the probability of age being less than 1 for a uniformly distributed machine

For a uniformly distributed machine whose age ranges from \(0\) to \(2\) years, the probability that its age is less than \(1\) year is simply the length of the interval between \(0\) and \(1\) (where age is less than 1) divided by the length of the entire interval \((0,2)\). Hence, the probability is: \(P(\text{Age} < 1) = \frac{1 - 0}{2 - 0} = \frac{1}{2}\)

Part (a): Convert the probability into percentage for a uniformly distributed machine

The probability we found is \(\frac{1}{2}\), which is equal to \(0.5\) in decimal form. To convert this probability into a percentage, multiply by \(100\%\): Percentage = \(0.5 * 100\% = 50\%\) Thus, for a uniformly distributed machine, it is in use less than one year old approximately \(50\%\) of the time.

Part (b): Determine the probability of age being less than 1 for an exponentially distributed machine

For an exponentially distributed machine with mean \(1\), we have to integrate the probability density function from \(0\) to \(1\) to find the probability that the age is less than \(1\) year. In this case, the probability density function is \(f(t) = e^{-t}\). The probability is then given by: \(P(\text{Age} < 1) = \int_0^1 e^{-t} dt\)

Part (b): Calculate the integral for an exponentially distributed machine

To calculate the integral, we find the antiderivative of the integrand, which is: \(F(t) = -e^{-t}\) Now, apply the limits of integration: \(P(\text{Age} < 1) = [-e^{-t}]_0^1 = -e^{-1} - (-e^{0}) = -e^{-1}+1\)

Part (b): Convert the probability into percentage for an exponentially distributed machine

The probability we found is \(-e^{-1}+1\), which is approximately \(0.6321\) in decimal form. To convert this probability into a percentage, multiply by \(100\%\): Percentage = \(0.6321 * 100\% \approx 63.21\%\) Thus, for an exponentially distributed machine, it is in use less than one year old approximately \(63.21\%\) of the time.