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A study by Bechtel et al., \(2009,\) described in the Archives of Environmental \& Occupational Health considered polycyclic aromatic hydrocarbons and immune system function in beef cattle. Some cattle were near major oil- and gas- producing areas of western Canada. The mean monthly exposure to PM1.0 (particulate matter that is \(<1 \mu \mathrm{m}\) in diameter) was approximately 7\. \(1 \mu \mathrm{g} / \mathrm{m}^{3}\) with standard deviation \(1.5 .\) Assume that the monthly exposure is normally distributed. (a) What is the probability of a monthly exposure greater than \(9 \mu g / m^{3} ?\) (b) What is the probability of a monthly exposure between 3 and \(8 \mu \mathrm{g} / \mathrm{m}^{3} ?\) (c) What is the monthly exposure level that is exceeded with probability \(0.05 ?\) (d) What value of mean monthly exposure is needed so that the probability of a monthly exposure more than \(9 \mu \mathrm{g} / \mathrm{m}^{3}\) is \(0.01 ?\)

Step by step solution

01

Find Z-score for Part (a)

To find the probability of a monthly exposure greater than 9 \( \mu g / m^{3} \), first calculate the Z-score using the formula:\[Z = \frac{X - \mu}{\sigma}\]Where \( X = 9 \), \( \mu = 7.1 \), and \( \sigma = 1.5 \).\[Z = \frac{9 - 7.1}{1.5} = \frac{1.9}{1.5} \approx 1.27\]

02

Probability for Part (a)

Using the Z-score obtained in Step 1, refer to the standard normal distribution table to find the probability. The table gives the cumulative probability up to \( Z = 1.27 \), which is approximately 0.8980. Therefore, the probability of a monthly exposure greater than 9 \( \mu g / m^{3} \) is:\[P(X > 9) = 1 - P(X \leq 9) = 1 - 0.8980 = 0.1020\]

03

Find Z-scores for Part (b)

First, find the Z-score for \( X = 3 \):\[Z_3 = \frac{3 - 7.1}{1.5} = \frac{-4.1}{1.5} \approx -2.73\]Then find the Z-score for \( X = 8 \):\[Z_8 = \frac{8 - 7.1}{1.5} = \frac{0.9}{1.5} \approx 0.60\]

04

Probability for Part (b)

Using the standard normal distribution table, find the cumulative probabilities for \( Z = -2.73 \) and \( Z = 0.60 \).\\( P(X \leq 3) \approx 0.0032 \) and \( P(X \leq 8) \approx 0.7257 \).\Thus, the probability of \( X \) between 3 and 8 is:\[P(3 < X < 8) = P(X \leq 8) - P(X \leq 3) = 0.7257 - 0.0032 = 0.7225\]

05

Find Monthly Exposure Level for Part (c)

For the monthly exposure level that is exceeded with probability 0.05, find the Z-score corresponding to 0.95 (since we're looking for \( P(X > x) = 0.05 \), i.e., \( P(X \leq x) = 0.95 \)).From the Z-table, \( Z = 1.645 \).Use the Z-score formula to find the value:\[1.645 = \frac{x - 7.1}{1.5} \Rightarrow x = 1.645 \times 1.5 + 7.1 = 9.568\]So, the monthly exposure level is approximately \( 9.57 \mu g / m^{3} \).

06

Mean Monthly Exposure for Part (d)

Given \( P(X > 9) = 0.01 \), find corresponding Z-score for \( P(X \leq 9) = 0.99 \). From the Z-table, \( Z = 2.33 \).Using the Z-score for \( X = 9 \), setting up the equation:\[2.33 = \frac{9 - \mu}{1.5} \Rightarrow \mu = 9 - 2.33 \times 1.5 = 5.505\]The mean monthly exposure needed is approximately \( 5.51 \mu g / m^{3} \).

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Key Concepts

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

Z-score

In the realm of statistics, a Z-score is a very useful number. It helps you understand where a particular data point stands in relation to the mean of a data set. When dealing with normal distribution, the Z-score or standard score tells you how many standard deviations the data point is from the mean.
To calculate the Z-score, use the formula: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
For example, if the mean monthly exposure is 7.1 and the standard deviation is 1.5, to find the Z-score for an exposure of 9, plug the numbers into the formula and get \( Z \approx 1.27 \). This means the exposure level of 9 \( \mu g / m^{3} \) is 1.27 standard deviations above the mean.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simpler terms, it tells us how spread out the values are in a dataset.
In a normal distribution, data is equally distributed around the mean, and the standard deviation helps to understand how far each data point is from that mean. A smaller standard deviation means data points are closer to the mean, while a larger one indicates they are spread out.
In our exercise, the standard deviation is \( 1.5 \), which means most of the exposure values are within 1.5 units away from the mean of \( 7.1 \mu g / m^{3} \). It allows us to calculate probabilities and understand variations in monthly exposure.

Cumulative Probability

Cumulative probability is a concept in probability that refers to the likelihood that a random variable is less than or equal to a particular value. It helps us understand the probability of observations falling below a certain point in a dataset.
When dealing with normal distribution, the cumulative probability up to a certain Z-score can be found using a standard normal distribution table. In the example, the cumulative probability for \( Z = 1.27 \) is approximately 0.8980. This tells us that there is an 89.80% chance of a monthly exposure level being less than or equal to 9 \( \mu g / m^{3} \).
To find the probability of an exposure greater than a certain value, simply subtract the cumulative probability from 1, as demonstrated in the original solution.

Probability Calculations

Probability calculations using Z-scores and the standard normal distribution table allow us to predict the likelihood of different outcomes. This involves several steps:

• Calculate the Z-score for the value you are interested in.
• Look up the cumulative probability for that Z-score from the standard normal distribution table.
• Use the cumulative probability to find the probability of your specific event.

For instance, to find the probability of monthly exposure being between two values, say from 3 to 8 \( \mu g / m^{3} \), you calculate the Z-scores for both values. Then, find their cumulative probabilities, and subtract one from the other.
These calculations help us make informed predictions based on data. They are especially useful in fields like environmental health, where understanding exposure levels to pollutants is crucial.