2. The number of accidents (A) at an intersection is counted for a 12-hour period. The number of accidents follows a Poisson distribution with a parameter value of 3.5. (Round to 3 decimal places and use a point-not a comma).Calculate the following:a. 𝑃(𝐴 = 1) Blank 1b. 𝑃(𝐴 < 5) Blank 2c. 𝑃(𝐴 > 0) Blank 3d. 𝑃(0 < 𝐴 ≤ 3) Blank 4e. The probability that there will be 4 accidents in a 24-hour period Blank 5f. The expected number of accidents per week. Blank 6g. Let 𝐵=number of hours in a 12-hour cycle that there are no accidents. Calculate the probability that for 8 hours in the 12-hour cycle there will be no accidents. Blank 7
Question
- The number of accidents (A) at an intersection is counted for a 12-hour period. The number of accidents follows a Poisson distribution with a parameter value of 3.5. (Round to 3 decimal places and use a point-not a comma).Calculate the following:a. 𝑃(𝐴 = 1) Blank 1b. 𝑃(𝐴 < 5) Blank 2c. 𝑃(𝐴 > 0) Blank 3d. 𝑃(0 < 𝐴 ≤ 3) Blank 4e. The probability that there will be 4 accidents in a 24-hour period Blank 5f. The expected number of accidents per week. Blank 6g. Let 𝐵=number of hours in a 12-hour cycle that there are no accidents. Calculate the probability that for 8 hours in the 12-hour cycle there will be no accidents. Blank 7
Solution
a. The probability of exactly one accident occurring in a 12-hour period can be calculated using the formula for the Poisson distribution:
P(A = 1) = λ^k * e^-λ / k!
where λ is the average rate of occurrence (3.5 in this case), k is the number of occurrences we're interested in (1 in this case), and e is the base of the natural logarithm (approximately 2.71828).
P(A = 1) = 3.5^1 * e^-3.5 / 1! = 0.105
So, Blank 1 = 0.105
b. The probability of fewer than five accidents occurring is the sum of the probabilities of 0, 1, 2, 3, and 4 accidents occurring.
P(A < 5) = P(A = 0) + P(A = 1) + P(A = 2) + P(A = 3) + P(A = 4)
You can calculate each of these probabilities using the same formula as above, and then add them together.
So, Blank 2 = P(A < 5)
c. The probability of more than zero accidents occurring is 1 minus the probability of zero accidents occurring.
P(A > 0) = 1 - P(A = 0)
So, Blank 3 = P(A > 0)
d. The probability of more than zero but less than or equal to three accidents occurring is the sum of the probabilities of 1, 2, and 3 accidents occurring.
P(0 < A ≤ 3) = P(A = 1) + P(A = 2) + P(A = 3)
So, Blank 4 = P(0 < A ≤ 3)
e. The probability of exactly four accidents occurring in a 24-hour period is calculated the same way as in part a, but with λ doubled (since the time period is twice as long).
So, Blank 5 = P(A = 4) for λ = 7
f. The expected number of accidents per week is simply 7 times the average rate of occurrence per day (since there are 7 days in a week).
So, Blank 6 = 7 * λ
g. The probability of no accidents occurring in an 8-hour period is calculated the same way as in part a, but with λ scaled down by 2/3 (since the time period is 2/3 as long).
So, Blank 7 = P(A = 0) for λ = 3.5 * 2/3
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