Problem 100 - Applied Statistics and Probability for Engineers

Question

Phoenix water is provided to approximately 1.4 million people who are served through more than 362,000 accounts (http:// phoenix.gov/WATER/wtrfacts.html). All accounts are metered and billed monthly. The probability that an account has an error in a month is \(0.001,\) and accounts can be assumed to be independent. (a) What are the mean and standard deviation of the number of account errors each month? (b) Approximate the probability of fewer than 350 errors in a month. (c) Approximate a value so that the probability that the number of errors exceeds this value is \(0.05 .\) (d) Approximate the probability of more than 400 errors per month in the next two months. Assume that results between months are independent.

Step-by-Step Solution

Answer

Mean: 362; SD: 19.02. (a) 25.46%. (b) ~393 errors. (c) 0.23%.

Step 1: Identify the Distribution

Given the large number of accounts (362,000) and small probability of a single account error (0.001), it's appropriate to use the Poisson distribution to model the number of account errors per month. This is because the Poisson distribution is well-suited for estimating the number of events that happen independently within a fixed interval of time or space, given a large number of trials with a small expected number of successes each.

Step 2: Calculate the Mean of the Poisson Distribution

For a Poisson distribution, the mean, \(\lambda\), is given by the product of the number of trials and the probability of success: \(\lambda = 362,000 \times 0.001 = 362\). Hence, the expected number of account errors per month is 362.

Step 3: Calculate the Standard Deviation of the Poisson Distribution

In a Poisson distribution, the standard deviation is the square root of the mean: \(\sigma = \sqrt{362} \approx 19.02\). Therefore, the standard deviation of the number of account errors each month is approximately 19.02.

Step 4: Approximate Probability for Fewer than 350 Errors

Use the normal approximation to the Poisson distribution, since \(\lambda = 362\) is large. The distribution can be approximated by a normal distribution with mean 362 and standard deviation 19.02. Calculate \(Z\) for 349.5 (using continuity correction): \(Z = \frac{349.5 - 362}{19.02} \approx -0.66\). Using the standard normal distribution table, \(P(Z < -0.66) \approx 0.2546\). Thus, the probability of fewer than 350 errors is approximately 25.46%.

Step 5: Find the Value Exceeding 5% Probability

To find the number of errors such that the probability of having more than that number is 5%, we need to find \(Z_{0.95}\), which is approximately 1.645. Solve for \(X\) using \(X - 362 = 1.645 \times 19.02\), which gives \(X \approx 393.30\). The probability of exceeding approximately 393 errors is 5%.

Step 6: Calculate Probability for More Than 400 Errors in Two Months

Assume the total errors over two months follows a Poisson distribution with mean 2 \(\times 362 = 724\). We want \(P(X > 800)\). First, calculate \(Z\) for 800.5: \(Z = \frac{800.5 - 724}{\sqrt{724}} \approx 2.84\). Then, find \(P(Z > 2.84)\), which is approximately 0.0023. Thus, the probability of more than 400 errors per month in the next two months is about 0.23%.

Key Concepts

Mean and Standard Deviation

In probability theory, the Poisson distribution is key when modeling the number of times an event occurs within a specified interval. It's especially useful when dealing with situations where events are rare but have a large number of opportunities to occur, like water meter errors in this instance. Here, the mean (\(\lambda\)) of the number of errors per month can be calculated as the product of the number of accounts and the probability of error per account. This results in \(\lambda = 362,000 \times 0.001 = 362\). This tells us that, on average, you can expect 362 errors monthly.

The standard deviation gives us an idea of how much the number of errors might vary from the mean. For Poisson distributions, the standard deviation is simply the square root of the mean: \(\sigma = \sqrt{362} \approx 19.02\). This means most months, the number of errors will be within 19.02 of 362.

Probability Approximation

When dealing with a substantially large mean like 362, approximating probabilities with the normal distribution can simplify calculations. This transformation from Poisson to normal is a handy technique when computing probabilities becomes cumbersome due to large numbers.

Here, the goal is to approximate the probability that fewer than 350 errors happen in a month. By using the normal distribution, which has the same mean and standard deviation as Poisson, we can streamline the process. Applying the normal approximation, a continuity correction is used for discrete data. Instead of calculating probability directly for 350, we calculate it for 349.5, thus using the better-fitting continuous model. The calculation turns into finding the standard normal variable \(Z\): \(Z = \frac{349.5 - 362}{19.02} \approx -0.66\). Using standard normal distribution tables reveals that the probability \(P(Z < -0.66)\) is roughly 25.46%.

Normal Approximation

Normal approximation is a powerful method often employed when examining probabilities of rare events across large sample sizes, especially within the context of Poisson distributions. For instance, to find the number of account errors where exceeding this is only 5% probable, one utilizes normal distribution properties.

Using the normal distribution, we seek a value such that 95% of observations fall below it—prevalent in inventory and risk management applications. The Z-score corresponding to 95% cumulative probability \(Z_{0.95}\) is approximately 1.645. With this, set up: \(X - 362 = 1.645 \times 19.02\), leading to approximately \(X \approx 393.30\). Thus, there is a 5% probability that more than 393 account errors occur in a month.

Error Probability Calculation

Estimating the likelihood of an event happening over multiple intervals can often involve compounding effects. Here, the task is to calculate the probability of more than 400 errors per month occurring in the next two months. Adding the means for the two months transforms this into a problem involving the Poisson distribution with a new mean \(2 \times 362 = 724\).

Interested in finding \(P(X > 800)\), we utilize the normal approximation once more. Calculating \(Z\) for 800.5, applying continuity correction becomes necessary: \(Z = \frac{800.5 - 724}{\sqrt{724}} \approx 2.84\). Looking up standard normal distribution tables, \(P(Z > 2.84)\) is approximately 0.23%. This helps conclude that the chance of surpassing 400 errors in both upcoming months is quite slim at just about 0.23%.