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The demand for water use in Phoenix in 2003 hit a high of about 442 million gallons per day on June 27 (http://phoenix.gov/WATER/wtrfacts.html). Water use in the summer is normally distributed with a mean of 310 million gallons per day and a standard deviation of 45 million gallons per day. City reservoirs have a combined storage capacity of nearly 350 million gallons. a. What is the probability that a day requires more water than is stored in city reservoirs? b. What reservoir capacity is needed so that the probability that it is exceeded is \(1 \% ?\) c. What amount of water use is exceeded with \(95 \%\) probability? d. Water is provided to approximately 1.4 million people. What is the mean daily consumption per person at which the probability that the demand exceeds the current reservoir capacity is \(1 \% ?\) Assume that the standard deviation of demand remains the same.

Step by step solution

01

Define the Random Variable

Let the random variable \( X \) represent the daily water usage in Phoenix, modeled by a normal distribution with a mean \( \mu = 310 \) million gallons and a standard deviation \( \sigma = 45 \) million gallons. Thus, \( X \sim N(310, 45^2) \).

02

Calculate Probability for Part (a)

For part (a), we need to find \( P(X > 350) \). Find the z-score: \[ z = \frac{350 - 310}{45} = \frac{40}{45} \approx 0.89 \] Use the standard normal distribution table to find \( P(Z > 0.89) \). The table gives \( P(Z < 0.89) \approx 0.8133 \), so \( P(Z > 0.89) = 1 - 0.8133 = 0.1867 \). Thus, \( P(X > 350) \approx 0.1867 \).

03

Calculate Reservoir Capacity for Part (b)

For part (b), we need to find the reservoir capacity \( C \) such that \( P(X > C) = 0.01 \). This implies \( P(X \leq C) = 0.99 \). The z-score for \( 0.99 \) is approximately \( 2.33 \). Solve for \( C \) using the z-score formula: \[ 2.33 = \frac{C - 310}{45} \Rightarrow C = 2.33 \times 45 + 310 = 414.85 \] Therefore, a capacity of approximately 415 million gallons is needed.

04

Calculate Water Use for Part (c)

For part (c), find the water use \( W \) such that \( P(X > W) = 0.95 \), which implies \( P(X \leq W) = 0.05 \). The z-score for \( 0.05 \) is approximately \( -1.645 \). Use the z-score formula: \[ -1.645 = \frac{W - 310}{45} \Rightarrow W = -1.645 \times 45 + 310 = 235.975 \] Therefore, water use of approximately 236 million gallons is exceeded with \( 95\% \) probability.

05

Calculate Daily Consumption Per Person for Part (d)

For part (d), let \( D \) be the daily consumption per person such that \( P(X > 350) = 0.01 \). We know the z-score for \( 0.99 \) is \( 2.33 \). Set up the equation: \[ 2.33 = \frac{N - 1.4D}{45} \] where \( N = 310 \text{ million gallons} \). \[ N - 1.4D = 2.33 \times 45 + 310 = 414.85 \] Solving for \( D \) gives: \[ 1.4D = 414.85 \Rightarrow D = \frac{414.85}{1.4} = 296.32 \] The mean daily consumption per person is approximately 296.32 gallons.

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Key Concepts

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

Probability Calculation

Probability calculation in a normal distribution helps determine the likelihood of a particular outcome. This exercise involves understanding the probability that daily water demand exceeds the city reservoir's capacity. Consider the example where the water use's mean is 310 million gallons, with a standard deviation of 45 million gallons. For instance, the probability that more water than 350 million gallons will be needed in a day involves finding the z-score:

• The z-score formula is: \( z = \frac{X - \mu}{\sigma} \), where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
• Using the given values, \( z = \frac{350 - 310}{45} \approx 0.89 \).

Next, consult a standard normal distribution table to interpret this z-score. The table helps us find the probability that the water required exceeds the mentioned capacity. Understanding these steps is crucial for accurate probability outcomes.

Z-Score

The z-score is an essential concept in statistics, particularly for understanding normal distributions. It measures the number of standard deviations a data point is from the mean. In the context of the water consumption exercise, calculating the z-score helps decide when reservoir capacity is likely to be surpassed.Using the z-score formula \( z = \frac{X - \mu}{\sigma} \), you can translate water consumption values into z-scores. This conversion makes it easier to determine probabilities using the standard normal distribution table. For instance:

• A z-score of 0.89 for exceeding 350 million gallons means about 0.187 probability, interpreting approximately 18.7% likelihood.
• Z-scores assist in strategic planning, ensuring city water supply aligns with potential demand surges.

Utilizing z-scores aids in statistical analyses beyond just water consumption, fostering insightful data interpretations across fields.

Reservoir Capacity

Reservoir capacity refers to the storage amount available to meet water demands. Finding optimal reservoir capacity depends on understanding and predicting water usage patterns, shaped here by the normal distribution with given parameters. For this exercise, determining the reservoir capacity where the probability of being exceeded is just 1% is crucial.

• The calculation involves setting \( P(X \leq C) = 0.99 \) and finding the corresponding z-score, which is approximately 2.33 for 99% probability.
• You then rearrange the z-score formula: \( C = \mu + z\sigma \).
• Using the values, find \( C = 310 + 2.33 imes 45 = 414.85 \), suggesting nearly 415 million gallons.

This approach ensures that reservoirs are sufficiently prepared against severe demand days, maintaining city resilience.

Daily Water Consumption

Daily water consumption per person offers vital insights into individual usage patterns and overall demand met by city reservoirs. In this example, Phoenix's water distribution must account for potential excess days. Given 1.4 million people rely on the city's water, precise measurements contribute to forecasting and resource management.When calculating average daily consumption per person, the equation becomes:

• Use the equation: \( 1.4 imes D = 414.85 \), derived from total city needs and supplies
• Rearranging, you find \( D = \frac{414.85}{1.4} \approx 296.32 \) gallons per person daily.

Understanding these metrics ensures efficient and effective water distribution, enhancing service quality, and infrastructure planning for metropolises.