lambda= B C D E F G \( P[X<3]= \) \( =P[X<=2]= \) \( \square \) \( 0.173578= \) POISSON.DIST(2,B1,TRUE) Example Arrivals at a busstop follow a Poisson distribution with an average of 4.5 every quarter of an hour. Calculate the probability of fewer than 3 arrivals in a quarter of an hour.

To solve the problem, we first note that when we say fewer than 3 arrivals, it is equivalent to finding the probability of 0, 1, or 2 arrivals. For a Poisson distribution with an average rate (λ) of 4.5 arrivals per quarter hour, the probability of k arrivals is given by the formula:   P[X = k] = (λ^k * e^(–λ)) / k! Thus, the probability of fewer than 3 arrivals (k = 0, 1, or 2) is:   P[X < 3] = P[X ≤ 2] = P[X = 0] + P[X = 1] + P[X = 2] Calculating each term: 1. For k = 0:   P[X = 0] = (4.5^0 * e^(–4.5)) / 0! = e^(–4.5) ≈ 0.011109 2. For k = 1:   P[X = 1] = (4.5^1 * e^(–4.5)) / 1! = 4.5 * e^(–4.5) ≈ 0.049994 3. For k = 2:   P[X = 2] = (4.5^2 * e^(–4.5)) / 2! = (20.25 * e^(–4.5)) / 2 ≈ 10.125 * e^(–4.5) ≈ 0.112475 Adding these probabilities:   P[X < 3] ≈ 0.011109 + 0.049994 + 0.112475 = 0.173578 In Excel, you can calculate this cumulative probability using the POISSON.DIST function:   =POISSON.DIST(2, 4.5, TRUE) This will return approximately 0.173578. Therefore, the probability of fewer than 3 arrivals in a quarter of an hour is approximately 17.36%.