Final answer:
The probability that an alkaline battery will fail before 185 days is approximately 0.457, or 45.7%. The probability that an alkaline battery will last beyond 2 years is approximately 0.228, or 22.8%.
Step-by-step explanation:
To answer part (a), we need to find the probability that an alkaline battery will fail before 185 days. Since the length of life of an alkaline battery follows an exponential distribution with an average life of 340 days, we can use the formula for the exponential distribution to calculate the probability. The formula is:
P(X < t) = 1 - e^(-t/μ)
Where X is the random variable representing the length of life of the battery, t is the time (in days) we are interested in, and μ is the average life of the battery. Plugging in the values, we have:
P(X < 185) = 1 - e^(-185/340)
Solving this equation gives us a probability of approximately 0.457, or 45.7%.
For part (b), we want to find the probability that an alkaline battery will last beyond 2 years. Since there are 365 days in a year, 2 years is equal to 730 days. Using the same formula as before, we have:
P(X > 730) = 1 - P(X < 730) = 1 - (1 - e^(-730/340))
Solving this equation gives us a probability of approximately 0.228, or 22.8%.