Final answer:
To find the probabilities for the lifetime of a certain electronic component with an exponential distribution and a mean of 2,000 hours, we can use the exponential distribution formula. The probability that the component will last at most 2,400 hours is 71.35%, the probability that it will last at least 1,600 hours is 44.93%, and the probability that it will last between 1,800 and 2,200 hours is 7.37%.
Step-by-step explanation:
To find the probabilities for the lifetime of the electronic component, we can use the exponential distribution. In this case, the mean of the distribution is 2,000 hours.
a) To find the probability that the component will last at most 2,400 hours, we need to calculate the cumulative distribution function (CDF) at 2,400 hours using the exponential distribution formula. The formula is P(X ≤ x) = 1 - e^(-x/μ), where x is the given value and μ is the mean. Plugging in the values, we get P(X ≤ 2,400) = 1 - e^(-2,400/2,000) = 1 - e^(-1.2) = 0.7135 or 71.35%.
b) To find the probability that the component will last at least 1,600 hours, we need to calculate the survival function (SF) at 1,600 hours using the exponential distribution formula. The formula is P(X > x) = e^(-x/μ), where x is the given value and μ is the mean. Plugging in the values, we get P(X > 1,600) = e^(-1,600/2,000) = e^(-0.8) = 0.4493 or 44.93%.
c) To find the probability that the component will last between 1,800 and 2,200 hours, we need to calculate the difference between the survival functions at those two points. P(1,800 ≤ X < 2,200) = P(X > 1,800) - P(X > 2,200) = e^(-1,800/2,000) - e^(-2,200/2,000) = e^(-0.9) - e^(-1.1) = 0.4066 - 0.3329 = 0.0737 or 7.37%.