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The article "Reliability of Domestic-Waste Biofilm Reactors" (J. of Envir. Engr., 1995: 785-790) suggests that substrate concentration \(\left(\mathrm{mg} / \mathrm{cm}^{3}\right)\) of influent to a reactor is normally distributed with \(\mu=.30\) and \(\sigma=.06\). a. What is the probability that the concentration exceeds \(.50 ?\) b. What is the probability that the concentration is at most \(.20\) ? c. How would you characterize the largest \(5 \%\) of all concentration values?

Short Answer

Expert verified
a. Approximately 0.0004. b. Approximately 0.0475. c. Values exceed approximately 0.3987 mg/cm³.

Step by step solution

01

Understand the Normal Distribution

We know that substrate concentration is normally distributed with a mean \( \mu = 0.30 \) and standard deviation \( \sigma = 0.06 \). We will use the properties of the normal distribution to calculate probabilities for various events. You can use the Z-score formula \( Z = \frac{X - \mu}{\sigma} \) to find out how many standard deviations a particular value \( X \) is from the mean.
02

Calculate probability of exceeding 0.50

To find \( P(X > 0.50) \), calculate the Z-score: \( Z = \frac{0.50 - 0.30}{0.06} = \frac{0.20}{0.06} \approx 3.33 \). Look up this Z-score in the standard normal distribution table or use a calculator to find \( P(Z > 3.33) \). The result is approximately 0.0004.
03

Calculate probability of at most 0.20

To find \( P(X \leq 0.20) \), calculate the Z-score: \( Z = \frac{0.20 - 0.30}{0.06} = \frac{-0.10}{0.06} \approx -1.67 \). Look up this Z-score in the standard normal distribution table or use a calculator to find \( P(Z \leq -1.67) \). The result is approximately 0.0475.
04

Characterize the largest 5% of concentrations

To find the cutoff for the largest 5% of data points, we need the Z-score corresponding to the 95th percentile of the normal distribution. This Z-score is approximately 1.645. Using the formula for Z-score: \( X = \mu + Z \times \sigma \), substitute \( \mu = 0.30 \), \( Z = 1.645 \), \( \sigma = 0.06 \). Thus, \( X = 0.30 + 1.645 \times 0.06 \approx 0.3987 \). So, the largest 5% of concentration values exceed approximately 0.3987 mg/cm³.

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Key Concepts

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

Probability Calculation
Probability calculations are essential in many applications, especially when dealing with data that follows a certain pattern or distribution. In this exercise, we looked at concentrations which follow a normal distribution. To calculate the probability of a concentration exceeding a specific value, like 0.50 mg/cm³, we first need to convert the concentration value into a Z-score. This is done using the formula:
  • Find the difference between the value in question and the mean (\(X - \mu\)).
  • Divide this difference by the standard deviation (\(\sigma\)).
This process helps tell us how far away any observed concentration value is from the average, measured in terms of standard deviations. By examining these probabilities, we can better understand the relative likelihoods of different scenarios in the dataset. This makes probability a valuable tool for evaluating risks and making informed decisions.
Z-score
The Z-score is a crucial statistic used to determine how an individual data point relates to the mean of a data set. It expresses this relationship in terms of standard deviations. Calculating a Z-score, especially when dealing with normal distribution, is a key step. You use the formula: \[ Z = \frac{X - \mu}{\sigma} \]where:
  • \(X\)is the data point we are evaluating.
  • \(\mu\)is the mean of the data set.
  • \(\sigma\)is the standard deviation.
A positive Z-score suggests that the data point is above the mean, while a negative Z-score indicates it is below the mean. For instance, a Z-score of 3.33, as calculated earlier for a concentration of 0.50, illustrates that the concentration is 3.33 standard deviations above the average concentration of 0.30. Understanding Z-scores is fundamental for interpreting which values are typical and which are outliers.
Standard Normal Distribution
The standard normal distribution is a normalized version of the normal distribution where the mean is 0 and the standard deviation is 1. This makes it handy because it allows us to compare different normal distributions by standardizing them. To use a standard normal distribution, we transform data points from any normal distribution to fit into this model using the Z-score, making different datasets directly comparable.
By using Z-scores and standardizing them, we can refer to tables or use statistical software to easily find probabilities for specific ranges, such as finding how rare a specific event is. In our example, once the Z-scores were calculated, they were used to look up the probability of exceeding \(3.33\)or being at most \(-1.67\) in standard normal distribution tables, leading to insights about how common or uncommon those concentration levels are.
Statistics for Engineering
In engineering, statistics play a critical role in problem-solving and decision-making. Engineering applications frequently leverage the normal distribution to model and analyze variances in processes, materials, or component behaviors. Understanding and applying concepts like probability, Z-scores, and the standard normal distribution are integral.
  • Engineers use these statistical tools to design systems and processes with appropriate safety margins.
  • They ensure product reliability by understanding and minimizing risks.
  • These statistical methodologies are crucial for quality control and optimizing performance.
In our example involving biofilm reactors, understanding the statistics behind substrate concentration helps engineers predict and manage variations in reactor performance. Thus, mastering these statistical concepts is essential for sound engineering practice.

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Most popular questions from this chapter

The article "A Model of Pedestrians" Waiting Times for Street Crossings at Signalized Intersections" (Transportation Research, 2013: 17-28) suggested that under some circumstances the distribution of waiting time \(X\) could be modeled with the following pdf: $$ f(x ; \theta, \tau)=\left\\{\begin{array}{cl} \frac{\theta}{\tau}(1-x / \tau)^{\theta-1} & 0 \leq x<\tau \\ 0 & \text { otherwise } \end{array}\right. $$ a. Graph \(f(x ; \theta, 80)\) for the three cases \(\theta=4,1\), and .5 (these graphs appear in the cited article) and comment on their shapes. b. Obtain the cumulative distribution function of \(X\). c. Obtain an expression for the median of the waiting time distribution. d. For the case \(\theta=4, \tau=80\), calculate \(P(50 \leq X \leq 70)\) without at this point doing any additional integration.

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