Examples of Exponential distribution

The exponential distribution is commonly used to model the time between events in a Poisson process. It is defined by a single parameter, $\lambda$ , which is the rate parameter. The probability density function (PDF) of the exponential distribution is:

$f(x; \lambda) = \lambda e^{-\lambda x}$

for $x \ge 0$ and $\lambda > 0$ .

Here are a few examples:

Example 1: Time Between Arrivals

Suppose the average number of customers arriving at a bank is 3 per hour. We want to find the probability that the time until the next customer arrives is less than 10 minutes.

Determine $\lambda$ :

Since there are 3 customers per hour, $\lambda = 3$ .

Convert the time to the same units as $\lambda$ :

10 minutes is $\frac{10}{60}$ hours, or $\frac{1}{6}$ hours.

Calculate the probability:

We need to find $P(X < \frac{1}{6})$ , where $X$ is the time until the next arrival.
The cumulative distribution function (CDF) of the exponential distribution is $F(x; \lambda) = 1 - e^{-\lambda x}$ .
Therefore, $P(X < \frac{1}{6}) = 1 - e^{-3 \cdot \frac{1}{6}} = 1 - e^{-0.5} \approx 1 - 0.6065 = 0.3935$ .

So, the probability that the next customer arrives within 10 minutes is approximately 0.3935, or 39.35%.

Example 2: System Failure Time

Suppose a machine has a failure rate of 0.02 failures per hour. We want to find the expected time until the next failure and the probability that the machine will operate for at least 50 hours without failure.

Determine $\lambda$ :

The failure rate is 0.02 per hour, so $latex\lambda = 0.02$.

Expected time until the next failure:

The expected value $E[X]$ of an exponential distribution is $\frac{1}{\lambda}$ .
Therefore, $E[X] = \frac{1}{0.02} = 50$ hours.

Calculate the probability:

We need to find $P(X > 50)$ .
The survival function is $P(X > x) = e^{-\lambda x}$ .
Therefore, $P(X > 50) = e^{-0.02 \cdot 50} = e^{-1} \approx 0.3679$ .

So, the expected time until the next failure is 50 hours, and the probability that the machine will operate for at least 50 hours without failure is approximately 0.3679, or 36.79%.

Example 3: Queue at a Toll Booth

Suppose cars arrive at a toll booth at an average rate of 5 cars per minute. We want to find the probability that the next car arrives in more than 2 minutes.

Determine $\lambda$ :

The average rate is 5 cars per minute, so $latex\lambda = 5$.

Calculate the probability:

We need to find $P(X > 2)$ , where $latexX$ is the time until the next car arrives.
The survival function is $P(X > x) = e^{-\lambda x}$ .
Therefore, $P(X > 2) = e^{-5 \cdot 2} = e^{-10} \approx 4.54 \times 10^{-5}$ .

So, the probability that the next car arrives in more than 2 minutes is approximately $4.54 \times 10^{-5}$ , or 0.00454%.

Example 4: Time to Failure of a Component

Suppose a particular type of electronic component has an average failure time of 200 hours. We want to find the probability that a component fails within the first 50 hours.

Determine $\lambda$ :

The average failure time is 200 hours, so $\lambda = \frac{1}{200}$ .

Calculate the probability:

We need to find $P(X < 50)$ , where $latexX$ is the failure time.
The CDF is $F(x; \lambda) = 1 - e^{-\lambda x}$ .
Therefore, $P(X < 50) = 1 - e^{-\frac{1}{200} \cdot 50} = 1 - e^{-0.25} \approx 1 - 0.7788 = 0.2212$ .

So, the probability that a component fails within the first 50 hours is approximately 0.2212, or 22.12%.

Example 5: Time Between Bus Arrivals

Suppose buses arrive at a bus stop every 15 minutes on average. We want to find the probability that the next bus arrives within the next 5 minutes.

Determine $\lambda$ :

The average arrival rate is $\frac{1}{15}$ per minute, so $\lambda = \frac{1}{15}$ .

Calculate the probability:

We need to find $P(X < 5)$ , where $X$ is the time until the next bus arrives.
The CDF is $F(x; \lambda) = 1 - e^{-\lambda x}$ .
Therefore, $P(X < 5) = 1 - e^{-\frac{1}{15} \cdot 5} = 1 - e^{-\frac{1}{3}} \approx 1 - 0.7165 = 0.2835$ .

So, the probability that the next bus arrives within the next 5 minutes is approximately 0.2835, or 28.35%.

Example 6: Time Until Call Center Receives a Call

Suppose a call center receives calls at a rate of 10 calls per hour. We want to find the expected time until the next call and the probability that the next call comes in less than 3 minutes.

Determine $\lambda$ :

The rate is 10 calls per hour, so $\lambda = 10$ .

Expected time until the next call:

The expected value $E[X]$ is $\frac{1}{\lambda}$ .
Therefore, $E[X] = \frac{1}{10}$ hours, or 6 minutes.

Calculate the probability:

We need to find $P(X < \frac{3}{60})$ hours.
The CDF is $F(x; \lambda) = 1 - e^{-\lambda x}$ .
Therefore, $P(X < \frac{3}{60}) = 1 - e^{-10 \cdot \frac{3}{60}} = 1 - e^{-0.5} \approx 1 - 0.6065 = 0.3935$ .

So, the expected time until the next call is 6 minutes, and the probability that the next call comes in less than 3 minutes is approximately 0.3935, or 39.35%.

Example 7: Time Until Next Defective Item

Suppose a manufacturing process produces defective items at an average rate of 1 defect per 100 items. We want to find the probability that the first defect occurs within the first 20 items.

Determine $\lambda$ :

The rate is 1 defect per 100 items, so $\lambda = \frac{1}{100}$ .

Calculate the probability:

We need to find $P(X < 20)$ , where $X$ is the number of items until the first defect.
The CDF is $F(x; \lambda) = 1 - e^{-\lambda x}$ .
Therefore, $P(X < 20) = 1 - e^{-\frac{1}{100} \cdot 20} = 1 - e^{-0.2} \approx 1 - 0.8187 = 0.1813$ .

So, the probability that the first defect occurs within the first 20 items is approximately 0.1813, or 18.13%.

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