Suppose that a die-hardly-ever battery has an exponential time-to-failure distribution with a mean of 48 months. at 60 months, the battery is still operating. what is the probability that the battery is going to last for the next 12 months?

Since the time to failure distribution is exponential, the distribution is F (n) = 1 – e ^ (-x/48)

The probability of failure between months 60 and 72 is the same as the probability that the battery will fail in 12 months, since the exponential distribution is memoryless F (12) = 0.221

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

take it another