Question

Cars arrive at a parking garage at a rate of 90 veh/hr according to the Poisson distribution. () In form of a table, write down the probability density and cumulative probabilities for the random variable Xrepresenting "the number of arrivals per minute forx -0 to 6, correct your answer to nearest 4 decimal places. P(X=x) F(x) P(Xsx) Find x such that there is at least 95% chance that the arrival rate is less than x vehicles per minutes. (ii) ii) What is (name of the probability distribution of the headway? Find the mean headway (in sec) and state the equation of the cumulative headway distribution iv Find the probability that a headway is less than or equal to 45 sec (v) The probability that a headway is longer than 2 min

Answer 1

(i)

$\lambda$ = 90 vehicles per hour = 90/60 vehicles per minute = 1.5 vehicles per minute.

Thus, X ~ Poisson($\lambda$ = 1.5)

P(X = x) = 1.5^x * exp(-1.5) / x!

x	P(X = x)	F(x) = P(X $\le$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) $\ge$ 0.95

From the table, in part (a), x = 4 such that F(x) $\ge$ 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 $\lambda$ = 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 $\le$ 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 $\le$ 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$\le$ 2)

=  exp(-1.5 * 2)

= 0.0498