Question

The manager of a gas station has observed that the times required by drivers to fill their car's tank and pay are quite variable. In fact, the times are exponentially distributed with a mean of 8 minutes. What is the probability that a car can complete the transaction in less than 5 minutes

Answer 1

Probability that a car can complete the transaction in less than 5 minutes is 0.465.

Explanation:

We are given that the manager of a gas station has observed that the times required by drivers to fill their car's tank and pay are quite variable. In fact, the times are exponentially distributed with a mean of 8 minutes.

Let X = times required by drivers to fill their car's tank and pay

The probability distribution function of exponential distribution is given by;

`f(x) = \lambda e^(-\lambda x) , x >0` where,
`\lambda` = parameter of distribution.

Now, the mean of exponential distribution is =
`(1)/(\lambda)` which is given to us as 8 minutes that means
`\lambda = (1)/(8)` .

So, X ~ Exp(
`\lambda = (1)/(8)` )

Also, we know that Cumulative distribution function (CDF) of Exponential distribution is given as;

`F(x) = P(X \leq x) = 1 - e^(-\lambda x)` , x > 0

Now, Probability that a car can complete the transaction in less than 5 minutes is given by = P(X < 5 min)

P(X < 5 min) =
`1 - e^{-(1)/(8) * 5}` {Using CDF}

= 1 - 0.5353 = 0.465

Therefore, probability that a car can complete the transaction in less than 5 minutes is 0.465.

Answer 2

The probability that a car driver can complete the transaction in less than 5 minutes is 0.4647.

Explanation:

Let X = times required by drivers to fill their car's tank and pay.

The average time required by drivers is, β = 8 minutes.

The random variable X is Exponentially distributed with parameter,
`\lambda=(1)/(\beta)=(1)/(8)=0.125`.

The probability distribution function of exponential distribution is:

`f_(X)(x)=\lambda e^(-\lambda x);\ x>0`

Compute the probability that a car driver can complete the transaction in less than 5 minutes as follows:

`P(X<5)=\int\limits^(5)_(0) {\lambda e^(-\lambda x)} \, dx\\=\lambda \int\limits^(5)_(0) {e^(-\lambda x)} \, dx\\=\lambda |(e^(-\lambda x))/(-lambda)|^(5)_(0) \\=|-e^(-\lambda x)|^(5)_(0)\\=-e^(-0.125* 5)+e^(-0.125* 0)\\=1-0.5353\\=0.4647`

Thus, the probability that a car driver can complete the transaction in less than 5 minutes is 0.4647.