Final answer:
To find the lifetime L that an electronic component is 60% certain to exceed, we solve the equation e^(-L) = 0.40. After finding L, the probability that at least one of five components fails before L years is calculated using 1 - (P(T > L))^5.
Step-by-step explanation:
The student is asking about finding the lifetime L which a typical electronic component is 60% certain to exceed given that the lifetime T is a continuous random variable with a probability density function f(t) = e−t; t≥0. To find L, we need to solve the equation P(T > L) = 0.60, which involves finding the complementary cumulative distribution function and setting up the integral 1 - ∫L∞ f(t) dt = 0.60. Once we find L, we can calculate the probability that at least one of the five components fails before L years using the probability 1 - P(all five exceed L), which is 1 - (P(T > L))^5.
With the exponential distribution, the probability that a component lasts longer than a certain time t is given by P(T > t) = e−t, so we set e−L = 0.40 (because P(T > L) = 0.60 implies 1 - P(T ≤ L) = 0.60). Solving for L gives L = −ln(0.40). After finding L, we can use the formula for the probability that at least one component fails before L years to get that probability.