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An auto mechanic recorded the number of car oil change services requested at his shop each day. He found that the average number of oil change services requested is 2.45 per day. Find the probability that on a randomly selected day, there are 5 car oil change services received. Use Excel to find the probability. - Round your answer to three decimal places.

Ask by Nichols Wilson. in the United States
Feb 03,2025

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Answer

The probability of receiving exactly 5 oil change services on a randomly selected day is **0.063**.

Solution

To find the probability that exactly 5 car oil change services are requested on a randomly selected day, we'll use the **Poisson distribution**. The Poisson distribution is appropriate here because we're dealing with the number of events (oil change services) happening in a fixed interval of time (one day), and these events occur with a known constant mean rate and independently of the time since the last event. ### Given: - **Average number of services per day (λ)**: 2.45 - **Number of services to find the probability for (k)**: 5 ### Poisson Probability Formula: \[ P(X = k) = \frac{{e^{-\lambda} \cdot \lambda^k}}{{k!}} \] ### Steps to Calculate Using Excel: 1. **Using the POISSON.DIST Function:** Excel provides the `POISSON.DIST` function to calculate Poisson probabilities. **Syntax:** ``` POISSON.DIST(x, mean, cumulative) ``` - `x`: The number of events (k) you want the probability for. - `mean`: The average number of events (λ). - `cumulative`: A logical value that determines the form of the function. Use `FALSE` for the probability mass function. **Formula for this problem:** ```excel =POISSON.DIST(5, 2.45, FALSE) ``` 2. **Calculating Manually:** If you prefer to understand the underlying calculation, here's how you'd compute it: \[ P(X = 5) = \frac{{e^{-2.45} \cdot 2.45^5}}{{5!}} \] - **Calculate \( e^{-2.45} \):** Approximately 0.086206 - **Calculate \( 2.45^5 \):** Approximately 88.1255 - **Calculate \( 5! \) (5 factorial):** 120 Plugging these values in: \[ P(X = 5) = \frac{{0.086206 \cdot 88.1255}}{{120}} \approx \frac{{7.6041}}{{120}} \approx 0.0634 \] ### Final Answer: Rounding to three decimal places, the probability that exactly 5 oil change services are requested on a randomly selected day is: **0.063** ### Example in Excel: Here's how you can input it in Excel: 1. Open Excel. 2. Click on a cell where you want the probability to appear. 3. Enter the formula: ```excel =POISSON.DIST(5, 2.45, FALSE) ``` 4. Press **Enter**. 5. The cell will display **0.063**. This means there's a **6.3%** chance of receiving exactly 5 oil change service requests on any given day.

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To find the probability of 5 car oil change services using Excel, you will use the POISSON.DIST function, which helps calculate the probability of a given number of events happening in a fixed interval of time, given a known average rate of occurrence. In this case, the average number of services (λ) is 2.45, and you want to find the probability of receiving exactly 5 services (k = 5). The formula you would use in an Excel cell is: `=POISSON.DIST(5, 2.45, FALSE)` This calculates the probability of getting exactly 5 services on a day when, on average, 2.45 services are expected. When you run this calculation, you’ll find that the probability rounds to approximately 0.042. So the probability of receiving exactly 5 oil changes on a randomly selected day is 0.042 (rounded to three decimal places, this would typically be represented as 0.042).

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