(i)
= 90 vehicles per hour = 90/60 vehicles per minute = 1.5 vehicles per minute.
Thus, X ~ Poisson( = 1.5)
P(X = x) = 1.5x * exp(-1.5) / x!
x | P(X = x) | F(x) = P(X x) |
0 | 0.2231 | 0.2231 |
1 | 0.3347 | 0.5578 |
2 | 0.2510 | 0.8088 |
3 | 0.1255 | 0.9343 |
4 | 0.0471 | 0.9814 |
5 | 0.0141 | 0.9955 |
6 | 0.0035 | 1 |
(ii)
We need to find x such that F(x) 0.95
From the table, in part (a), x = 4 such that F(x) 0.95
(iii)
We know that the time interval between Poisson processes follow exponential distribution.
The headway (time interval between cars) will follow exponential distribution with parameter = 1.5 per minute
Mean Headway = 1/ 1.5 = 0.6666667 minutes = 0.6666667 * 60 seconds = 40 Seconds
Cumulative headway distribution of exponential distribution is,
P(T t) = F(t) = 1 - exp(-1.5t)
(iv)
Probability that the headway is less than or equal to 45 sec = Probability that the headway is less than or equal to 45/60 min
P(T 45/60) = 1 - exp(-1.5 * 45/60)
= 0.6753
(v)
Probability that the headway is longer than 2 min = P(T > 2) = 1 - P(T 2)
= exp(-1.5 * 2)
= 0.0498
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