Final answer:
The probability density function for the lifespan of a two-component system operating in series, each with an exponential lifespan distribution, is f_X(x) = 1 - e^(-2x/100).
Step-by-step explanation:
The question is concerned with finding the probability density function (pdf) for the length of life of a system composed of two electronic components with independent and identical exponential distribution functions, operating in series. To determine the density function for X, the length of life of the system, we need to calculate the minimum of the two independent lifespan distributions Y1 and Y2. Since the lifespan of each component is exponentially distributed with f(y) = (1/100)e-y/100, we find that the density function for the system is:
f_X(x) = P(X > x) = P(Y1 > x and Y2 > x) = P(Y1 > x)P(Y2 > x) = [1 - P(Y1 ≤ x)][1 - P(Y2 ≤ x)],
Which simplifies to:
f_X(x) = [1 - (1 - e-x/100)][1 - (1 - e-x/100)] = [e-x/100][e-x/100] = e-2x/100. Therefore, the cumulative distribution function for X will be F_X(x) = 1 - e-2x/100.
The final answer: The probability density function for the length of life of a system composed of two independent and identically distributed exponential components in series is f_X(x) = 1 - e-2x/100.