Math Problem Statement

A new automated production process averages 1.5 breakdowns per day. Because of the cost associated with a breakdown, management is concerned about the possibility of having three or more breakdowns during a day. Assume that breakdowns occur randomly, that the probability of a breakdown is the same for any two time intervals of equal length, and that breakdowns in one period are independent of breakdowns in other periods. What is the probability of having three or more breakdowns during a day? Using excel

Solution

This problem is based on the Poisson distribution, which is used to model the number of events (breakdowns in this case) occurring within a fixed interval of time, where events happen independently and at a constant rate.

The Poisson distribution is defined by the formula:

P(X=k)=eλλkk!P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}

Where:

  • λ\lambda is the average number of events (breakdowns) in a given time period.
  • kk is the number of events.
  • ee is Euler's number (approximately 2.71828).

In this case:

  • λ=1.5\lambda = 1.5 (since the process averages 1.5 breakdowns per day).

We want to find the probability of 3 or more breakdowns. This can be calculated by first finding the probabilities of having 0, 1, and 2 breakdowns and then subtracting this sum from 1 (because the total probability must equal 1).

Steps to Calculate in Excel:

  1. Calculate probabilities for 0, 1, and 2 breakdowns:

    • Use the Poisson formula in Excel: =POISSON.DIST(k, λ, FALSE)
      • For 0 breakdowns: =POISSON.DIST(0, 1.5, FALSE)
      • For 1 breakdown: =POISSON.DIST(1, 1.5, FALSE)
      • For 2 breakdowns: =POISSON.DIST(2, 1.5, FALSE)
  2. Add the probabilities for 0, 1, and 2 breakdowns.

  3. Subtract the sum from 1 to get the probability of 3 or more breakdowns: P(X3)=1P(X=0)P(X=1)P(X=2)P(X \geq 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)

Example in Excel:

  1. For 0 breakdowns:
    =POISSON.DIST(0, 1.5, FALSE) → 0.2231

  2. For 1 breakdown:
    =POISSON.DIST(1, 1.5, FALSE) → 0.3346

  3. For 2 breakdowns:
    =POISSON.DIST(2, 1.5, FALSE) → 0.2510

  4. Total for 0, 1, and 2 breakdowns: P(X=0 or 1 or 2)=0.2231+0.3346+0.2510=0.8087P(X = 0 \text{ or } 1 \text{ or } 2) = 0.2231 + 0.3346 + 0.2510 = 0.8087

  5. The probability of 3 or more breakdowns: P(X3)=10.8087=0.1913P(X \geq 3) = 1 - 0.8087 = 0.1913

Thus, the probability of having three or more breakdowns during a day is approximately 0.1913, or 19.13%.

Tip for Excel:

When working with Poisson probabilities, you can also use the cumulative version to directly compute the probability of 2 or fewer breakdowns:
=POISSON.DIST(2, 1.5, TRUE) and then subtract this from 1.


Would you like more detailed steps or clarifications on the process?

Here are 5 related questions:

  1. What is the probability of exactly 3 breakdowns in a day?
  2. How would the probability change if the average breakdowns increased to 2 per day?
  3. Can you calculate the probability of fewer than 2 breakdowns using the cumulative function?
  4. How would you adjust this method to calculate the probability over a week instead of a day?
  5. Can you use a similar method to model rare events in different contexts, like system failures?

Tip: When using the Poisson distribution, the "cumulative" option in Excel saves time by avoiding manual summation.

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Math Problem Analysis

Mathematical Concepts

Statistics
Poisson Distribution
Probability

Formulas

P(X = k) = (e^(-λ) * λ^k) / k!
P(X ≥ 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)

Theorems

Poisson Theorem

Suitable Grade Level

College/University Level (Statistics or Probability Course)