91Ó°ÊÓ

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

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,

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

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 \(\left[ {\frac{1}{{{n^2}}} + \frac{1}{{{{\left( {n - 1} \right)}^2}}}} \right]{\mu ^2}\).