Question

Consider again the electronic system described in Exercise 10, but suppose now that the system will continue to operate until two components have failed. Determine the mean and the variance of the length of time until the system fails.

Short answer

The mean of the length of time until the system fails is\(\left[ {\frac{1}{n} + \frac{1}{{\left( {n - 1} \right)}}} \right]\mu \).

The variance of the time until the system fails is \(\left[ {\frac{1}{{{n^2}}} + \frac{1}{{{{\left( {n - 1} \right)}^2}}}} \right]{\mu ^2}\).

Step-by-step solution

Given information

An electronic system contains n similar components that function independently of each other.

Calculating the mean of the length of time until the system fails

The length of time\({Y_1}\)until one component fails has exponential distribution with the parameter\(n\beta \).

Then, the expected length will be,

\(E\left( {{Y_1}} \right) = \frac{1}{{n\beta }}\).

The additional length of time\({Y_2}\)until a second component fails has the exponential distribution with the parameter\(\left( {n - 1} \right)\beta \).

Then, the expected length will be,

\(E\left( {{Y_2}} \right) = \frac{1}{{\left[ {\left( {n - 1} \right)\beta } \right]}}\)

The length of the time until the system fails will be \({Y_1} + {Y_2}\)

Therefore, the expected length will be,

\(\begin{aligned}{}E\left( {{Y_1} + {Y_2}} \right) &= \frac{1}{{n\beta }} + \frac{1}{{{{\left[ {\left( {n - 1} \right)\beta } \right]}^2}}}\\ &= \left[ {\frac{1}{n} + \frac{1}{{\left( {n - 1} \right)}}} \right]\mu \end{aligned}\)

Calculating the variance of the length of time until the system fails

Since, the variables\({Y_1}\)and\({Y_2}\)are independent.

Therefore,

\(\begin{aligned}{}Var\left( {{Y_1} + {Y_2}} \right) = Var\left( {{Y_1}} \right) + Var\left( {{Y_2}} \right)\\ &= \frac{1}{{{{\left( {n\beta } \right)}^2}}} + \frac{1}{{{{\left[ {\left( {n - 1} \right)\beta } \right]}^2}}}\\& = \left[ {\frac{1}{{{n^2}}} + \frac{1}{{{{\left( {n - 1} \right)}^2}}}} \right]{\mu ^2}\end{aligned}\)

The mean of the length of time until the system fails is\(\left[ {\frac{1}{n} + \frac{1}{{\left( {n - 1} \right)}}} \right]\mu \).

The variance of length of time until the system fails is \(\left[ {\frac{1}{{{n^2}}} + \frac{1}{{{{\left( {n - 1} \right)}^2}}}} \right]{\mu ^2}\).