Answer:
Appropriate warranty period = 4 hours
Step-by-step explanation:
According to the scenario, computation of the given data are as follows:
Here, mean = 10,000 = 1/λ
Let appropriate warranty period = t
Hence, 0.0004 = 1-e^-λt
0.0004 = 1-e^(t/10,000)
e^(t/10,000) = 0.9996
Taking ln on both the sides and then solving,
t/10,000 = 0.0004
t = 4 hours
Hence, Appropriate warranty period = 4 hours