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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 is 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 pet month in the next two months Assume that results between months are independent.

Step by step solution

01

Mean of Monthly Errors

To find the mean (\(\mu\)) of the number of account errors each month, use the formula for the mean of a binomial distribution, which is \(\mu = np\). Here, \(n = 362,000\) (the number of accounts) and \(p = 0.001\) (the probability of an error). Thus, \(\mu = 362,000 \times 0.001 = 362\).

02

Standard Deviation of Monthly Errors

The standard deviation (\(\sigma\)) of a binomial distribution is calculated using \(\sigma = \sqrt{np(1-p)}\). Given \(n = 362,000\) and \(p = 0.001\), compute the standard deviation as \(\sigma = \sqrt{362,000 \times 0.001 \times 0.999} = \sqrt{361.638} \approx 19.01\).

03

Probability of Fewer than 350 Errors

Use the normal approximation of the binomial distribution with continuity correction to find this probability. The variable is approximately normally distributed with \(\mu = 362\) and \(\sigma \approx 19.01\). Calculate \(P(X < 350)\) using the z-score: \(z = \frac{349.5 - 362}{19.01} \approx -0.66\). Check the standard normal distribution table to find \(P(Z < -0.66) \approx 0.2546\).

04

Find Value for 0.05 Probability Exceedance

To find a value \(x\) so that \(P(X > x) = 0.05\), find \(P(Z < z) = 0.95\). Use a standard normal table to find \(z \approx 1.645\). Solve for \(x\) using \(x = \mu + z\sigma\): \(x = 362 + 1.645 \times 19.01 \approx 393.30\). Thus, \(P(X > 393) \approx 0.05\).

05

Probability of More than 400 Errors Over Two Months

For two months, the mean \(\mu_{2}\) becomes twice the monthly mean: \(362 \times 2 = 724\), and the standard deviation \(\sigma_{2} = \sqrt{2} \times \sigma = \sqrt{2} \times 19.01 \approx 26.88\). We need \(P(Y > 400 \times 2 = 800)\), calculate \(z = \frac{800 - 724}{26.88}\) which gives \(z \approx 2.83\). Using the standard normal distribution, \(P(Z > 2.83) \approx 0.0023\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability

Probability is the likelihood or chance of an event occurring. In our exercise, we have a probability of 0.001 for an account to have an error in a month.

• Probability can range from 0 (no chance) to 1 (certainty).
• Calculating probability helps in predicting future outcomes based on the likelihood of events.

In statistical terms, probability is often used to determine the expected number of occurrences, such as errors, in large sets of data. Here, knowing the probability allows us to calculate the average number of errors expected and their distribution.

Binomial Distribution

Binomial Distribution is a common statistical distribution used to model scenarios where there are two possible outcomes, such as success or failure, yes or no, and error or no error.

• This distribution is defined by two parameters: the number of trials () and the probability of success (p).
• For our water account example, each account represents a trial, and an error is considered a success (though undesirable).

The mean of a binomial distribution is calculated by the product of the number of trials and the probability (\(\mu = np\)). We also use binomial distribution to find the variance and standard deviation, which represent the spread of the distribution.

Normal Approximation

Normal Approximation is a statistical method used when dealing with a binomial distribution with a large number of trials and a moderate probability of success.

• This is especially useful because calculating probabilities directly from the binomial distribution can be complex when the sample size is large.
• For our water billing example, since there are 362,000 accounts, the binomial distribution can be approximated by a normal distribution, which simplifies calculations.

We use the normal approximation with the calculated mean and standard deviation of the binomial distribution. This involves using z-scores, which help determine how far off a particular value is from the mean in terms of standard deviations.

Standard Deviation

Standard Deviation is a measure of the dispersion or spread of a set of values. It indicates how much the values in a data set deviate from the mean.

• In a binomial distribution, the standard deviation can be computed using the formula \(\sigma = \sqrt{np(1-p)}\).
• It tells us how spread out the errors are each month from the average number of errors.

In the context of our exercise, finding the standard deviation helps us to quantify the certainty or uncertainty in the number of monthly account errors, giving insights about the consistency of these errors over time.