We were given the following formulas in the previous section, to calculate the probability of times between events:

• [latex]P(\text{at most or less than}) = P(X \le x) = 1 - e^{-\lambda x}[/latex]
• [latex]P(\text{at least or more than}) = P(X \ge x) = e^{-\lambda x}[/latex]

We will practice using these formulas in the sections below.

Calculating the probability of AT Most (EXAMPLE)

Let us better understand the Exponential Distribution formula by working through an example.

Example 37.1

Problem Setup: On a busy Friday evening, the time between customer arrivals at a supermarket check-out counter follows an Exponential distribution. On average, a customer arrives every 30 seconds.

Question: What is the probability a customer will arrive in the next minute at the check-out?

Written solution: Let us use the ‘at most’ formula and follow the steps below:

• Formula: [latex]P(\text{at most or less than}) = P(X \le x) = 1 - e^{-\lambda x}[/latex]

• Time units: We will use minutes as the time units.
• Lambda: We convert [latex]\lambda[/latex] to the number of events per minute: [latex]\lambda = \frac{1}{30 \text{ seconds}} \times \frac{60 \text{ seconds}}{1 \text{ minute}}=2 \text{ } \frac{\text{customers}}{\text{per minute}}[/latex]
• X-values: We want the probability of the next customer arriving within the next minute. This gives: [latex]P(x \le 1)[/latex].
• e: The other letter, [latex]e[/latex], is a constant. It is the base of the natural log and equal 2.71828.
• Answer: Plugging all of this into the formula [latex]P(X \le x) = 1 - e^{-\lambda x}[/latex] gives:
  [latex]\begin{align} P(X \le 1) &= 1 - e^{-2 \times 1} \\ &= 1 - e^{-2} \\ &= 1 - 0.1353 \\ &= 0.8647 \end{align}[/latex]

Conclusion: There is an 86.47% chance that a customer will arrive in the next minute at the check-out.

Calculating the probability of At least (Exercise)

Next, we will use the formulas to solve for the probability of at least a certain amount of time.

Example 37.2.1

Problem Setup: The times it takes call center specialist to resolve incoming calls to their call center:

• Follow an exponential distribution.
• Have an average of 2.5 minutes.

Question: Specialists will get a warning at 5 minutes to try to wrap up the call (if they can). What is the probability that the specialist receives that 5 minute warning on a call?

You Try: Fill in the values to answer the question in the exercise below:

Answer? Use the values from the above exercise to answer the question. Input your answer in the exercise below:

Applying the Memoryless Property (Video)

Let us now explore the memoryless property of the exponential distribution in the next example.

Example 37.2.2

Problem Setup: We will revisit our previous example…

• The times it takes call center specialist to resolve incoming calls follow an exponential distribution.
• On average, it takes the specialist 2.5 minutes to resolve a call.
• After 5 minutes of being on a call, a warning pops up on the specialist’s computer urging them to wrap up the phone call.

Question: What is the probability, after receiving the warning, that it takes at least another 5 minutes to wrap up the call?

Solution: Click here to download the written solutions shown in the video below:

Conclusion: There is 13.53% chance that the call will take at least another 5 minutes to resolve.

Key Takeaways (EXERCISE)

Your Own Notes (EXERCISE)

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