The number of automobiles entering a mountain tunnel per 2-minute period has a P

Question

The number of automobiles entering a mountain tunnel per 2-minute period has a Poisson distribution with mean one. An excessive number of cars entering the tunnel during a brief period of time produces a hazardous situation.

a) Find the probability that the number of autos entering the tunnel during a two-minute period exceeds three.

b) What is the probabilty that the number of autos entering the tunnel during a six-miniute period is less than 4?

c) Assume that the tunnel is observed during ten 2-minute intervals, thus giving independent obbservations Y1,Y2,...,Y10, on the Poisson random variable. Find the probabilty that Y>3 during at least one of the ten 2-minute interval. (Y= the number of autos during a 2-minute interval).

Explanation / Answer

when a random variable X follows a poisson distribution with parameter lambda (L), it means that X is a count of the number of occurrences of an "event" during a time interval of a certain length, where L is the expected(average) number of "events" that should occur during that particular time interval.

the probability MASS function for a poisson distribution with parameter L is...

P(X = n) = [(L^n)*(e^-L)]/(n!)

In words, this is the probability that exactly n "events" occur during a time interval such that the expected number of "events" for a time interval of that size is L. the "events" for this problem are automobiles entering the tunnel.

a.) we want to find P(X > 2). what is the lamda for this problem? well, the time interval under consideration for this specific problem is a 2-minute period. so L for this specific problem is the expected number of "events" (automobiles entering the tunnel) for a time interval of 2 minutes. according to the initial problem statement, "The average number of automobiles entering a mountain tunnel per 2-minute period is 1." so L=1. ok so to find P(X > 2), we would have to find P(X = 3) + P(X = 4) + P(X = 5) + ...... we would have to calculate an infinite number of probabilities and add them up! we cannot do this, so we use the fact that the probabilities for all different values of X must add up to 1: P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) +..... = 1
so to get P(X = 3) + P(X = 4) + P(X = 5) + ......, we will just take P(X = 0) + P(X = 1) + P(X = 2) and subtract it from 1. now we only have to calculate 3 numbers (as opposed to infinity) and subtract the result from 1!

so we get 1 - (0.3679 + 0.3679 + 0.1839) = 0.0803

b.) for this specific problem, the time interval is 6 minutes. so what is lambda for this problem? so L for this problem is the expected number of "events" for a time interval of 6 minutes. according to the initial problem statement, we have 1 automobile per 2 minutes, so that means 3 automobiles per 6 minutes. so L=3. now you have to find P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).

so we get 0.0498 + 0.1494 + 0.2240 = 0.4232

c.) this problem is about a different distribution than the poisson distribution. it is about the exponential distribution. much like the poisson distribution, the exponential distribution has a rate parameter lambda. when a random variable X follows an exponential distribution with parameter lambda (L), it means that X is a count of the amount of time before the first "event" (the first automobile enters the tunnel). unlike the poisson distribution, the exponential distribution is for continuous random variables, since X is a count of the amount of time for something, and time is a continous value (ie. you can have 2 seconds, 2.7838 seconds, 3.21 seconds, 10 seconds, etc.). in the previous problems, we were dealing with a discrete random variable X since was X the count of the number of times something happened and so X could only take the integer values 0, 1, 2, 3, etc. while discrete probability distributions generally have associated probability MASS functions, continuous probability distributions have probability DENSITY functions.

the probability DENSITY function for an exponential distribution with parameter L is...

f(x) = L*[e^(-L*x)] for x >= 0 , and f(x) = 0 for x < 0

unlike for probabiltiy MASS funtions, the values of probability DENSITY functions are NOT actual probabilities. for example, f(3) is NOT the probability that the first "event" occurs at a time of 3 time units (seconds, minutes, or whatever time unit you are using). for any continuous random variable X, the probability that X is equal to an exact value such as X=3 is 0. in other words P(X = 3) = 0. for continuous random variables such as X, you generally don't find the probability that X is equal to some exact value (becuase that probability is immediately known to be 0). instead you find the probability that X falls within a certain range of values such as X>2. so P(X >2) is the probability that the first "event" happens after 2 time units (minutes in our problem). the way to calculate the probability is to find the area under the curve or intergral of the probability DENSITY function from 2 to infinity.

in problem c), the distribution parameter is given as L = 1. the probability that no car enters the tunnel during the first 4 minutes is equivalent to the probability that the first "event" (car entering tunnel) occurs after 4 minutes, which is just P(X > 4). to get that, you could either take the integral of f(x) (the probability DENSITY function) from 4 to infinity, or you could take the integral of f(x) from 0 to 4 and subtract it from 1. to avoid infinities, let's use the second option.

since L=1,
f(x) = L*[e^(-L*x)] = e^(-x)

then do INTEGRAL FROM 0 TO 4: e^(-x) dx. the antiderivative of e^(-x) is just -e^(-x). to solve an integral you just plug in the limits of integrat

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