a. To find the probability that 5 traffic tickets will be issued at the intersection next month, we can use the Poisson distribution formula:
P(X = x) = (e^(-λ) * λ^x) / x!
where λ is the average number of events per unit of time or space. In this case, λ = 5.4 and x = 5. Plugging in these values, we get:
P(X = 5) = (e^(-5.4) * 5.4^5) / 5! = 0.169
Therefore, the probability that 5 traffic tickets will be issued at the intersection next month is 0.169 or about 17%.
b. To find the probability that 3 or fewer traffic tickets will be issued at the intersection next month, we need to find the cumulative probability up to 3:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Using the Poisson distribution formula as before, we get:
P(X ≤ 3) = (e^(-5.4) * 5.4^0) / 0! + (e^(-5.4) * 5.4^1) / 1! + (e^(-5.4) * 5.4^2) / 2! + (e^(-5.4) * 5.4^3) / 3!
P(X ≤ 3) = 0.095 + 0.258 + 0.350 + 0.266 = 0.969
Therefore, the probability that 3 or fewer traffic tickets will be issued at the intersection next month is 0.969 or about 97%.
c. To find the probability that more than 6 traffic tickets will be issued at the intersection next month, we can use the complement rule:
P(X > 6) = 1 - P(X ≤ 6)
We have already calculated P(X ≤ 3) in part b, so we can use the Poisson distribution formula to find P(X = 4), P(X = 5), and P(X = 6) and then add them up:
P(X = 4) = (e^(-5.4) * 5.4^4) / 4! = 0.136
P(X = 5) = (e^(-5.4) * 5.4^5) / 5! = 0.226
P(X = 6) = (e^(-5.4) * 5.4^6) / 6! = 0.237
P(X ≤ 6) = 0.095 + 0.258 + 0.350 + 0.266 + 0.136 + 0.226 + 0.237 = 1
Therefore, P(X > 6) = 1 - 1 = 0
Therefore, the probability that more than 6 traffic tickets will be issued at the intersection next month is 0 or 0%.