Final answer:
To calculate the probability of at least 3 lost-time accidents occurring over a period of 10 days, we can use the Poisson distribution. Additionally, we can use the Poisson distribution to find the probability of at most two accidents occurring in any given week. We can also use the geometric distribution to find the probability of there being at least two weeks between any two accidents.
Step-by-step explanation:
To calculate the probability that the number of lost-time accidents occurring over a period of 10 days will be at least 3, we can use the Poisson distribution. The Poisson distribution is often used to model the number of events that occur within a specified period of time. In this case, we know that the mean rate of lost-time accidents is 0.5 per day. The formula for the Poisson distribution is P(x;λ) = (e^(-λ) * λ^x) / x!, where x is the number of events, λ is the mean rate, e is Euler's number (approximately 2.71828), and x! is the factorial of x.
- To find the probability of at most two accidents occurring in any given week, we can sum the probabilities of 0, 1, and 2 accidents. Using the Poisson distribution formula, we can calculate P(0;3), P(1;3), and P(2;3) and add them together.
- To find the probability of there being at least two weeks between any two accidents, we can use the geometric distribution. The geometric distribution models the number of trials needed to achieve the first success. In this case, a success is defined as an accident occurring. The formula for the geometric distribution is P(X=k) = (1-p)^(k-1) * p, where X is a geometric random variable, k is the number of trials, and p is the probability of success.