Final answer:
To calculate the probability of at most 2 accidents daily, we use the Poisson distribution formula. By substituting the values and evaluating the expression, the probability is approximately 0.676. To find the probability of more than 5 accidents in a 5-day period, we calculate the complement of the probability of at most 5 accidents. Evaluating the expression, the probability is approximately 0.0039.
Step-by-step explanation:
a. Calculate the probability that at most two accidents occur in any given week:
Since the average number of daily accidents is 2, we can use the Poisson distribution to calculate the probability of at most 3 accidents daily. The formula for the Poisson distribution is:
P(X = k) = (e^(-λ) * λ^k) / k!
Where X is the random variable (number of accidents), λ is the average number of accidents (2), and k is the number of accidents we're interested in calculating the probability for.
To find the probability of at most 3 accidents daily, we can calculate:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Substituting the values into the formula:
P(X ≤ 3) = (e^(-2) * 2^0) / 0! + (e^(-2) * 2^1) / 1! + (e^(-2) * 2^2) / 2! + (e^(-2) * 2^3) / 3!
After evaluating the expression, the probability is approximately 0.676
b. Find the probability of more than 5 accidents in a 5-day period:
Since the average number of daily accidents is 2, the average number of accidents in a 5-day period would be 2 * 5 = 10. We can use the Poisson distribution again to calculate the probability of more than 5 accidents in a 5-day period.
To find this probability, we can calculate the complement of the probability of at most 5 accidents:
P(X > 5) = 1 - P(X ≤ 5) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5))
Using the same formula and substituting the values, we can evaluate the expression to find that the probability is approximately 0.0039.