91Ó°ÊÓ

Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential36. Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential is well known. The is well known. The paper "Effects of 2,4-D Formulation and Quinclorac on Spray Droplet Size and Deposition" (Weed Technology, 2005: 1030-1036) investigated the effects of herbicide formulation on spray atomization. A figure in the paper suggested the normal distribution with mean \(1050 \mu \mathrm{m}\) and standard deviation \(150 \mu \mathrm{m}\) was a reasonable model for droplet size for water (the "control treatment") sprayed through a \(760 \mathrm{ml} / \mathrm{min}\) nozzle. a. What is the probability that the size of a single droplet is less than \(1500 \mu \mathrm{m}\) ? At least \(1000 \mu \mathrm{m}\) ? b. What is the probability that the size of a single droplet is between 1000 and \(1500 \mu \mathrm{m}\) ? c. How would you characterize the smallest \(2 \%\) of all droplets? d. If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds \(1500 \mu \mathrm{m}\) ?

Step by step solution

01

Understanding the Normal Distribution

The problem suggests that the droplet size follows a normal distribution with mean \( \mu = 1050 \mu m \) and standard deviation \( \sigma = 150 \mu m \). We will use this information to calculate probabilities using the properties of the normal distribution.

02

Calculating Probability for Less Than 1500 µm

To find the probability that a droplet size is less than \( 1500 \mu m \), we compute the standard normal variable \( z \) using the formula \( z = \frac{x - \mu}{\sigma} \). Substitute \( x = 1500 \), \( \mu = 1050 \), and \( \sigma = 150 \): \( z = \frac{1500 - 1050}{150} = 3 \). Consult a standard normal distribution table or use a calculator to find \( P(Z < 3) \), which is about 0.9987.

03

Calculating Probability for At Least 1000 µm

For the probability that a droplet is at least \( 1000 \mu m \), we find \( P(X \geq 1000) \). Calculate \( z = \frac{1000 - 1050}{150} = -0.3333 \). Find \( P(Z > -0.3333) \), which equals \( 1 - P(Z < -0.3333) = 1 - 0.3707 = 0.6293 \).

04

Probability for the Range Between 1000 and 1500 µm

To find the probability that the droplet size is between \( 1000 \mu m \) and \( 1500 \mu m \), calculate \( P(1000 < X < 1500) = P(Z < 3) - P(Z < -0.3333) \). From earlier calculations, \( 0.9987 - 0.3707 = 0.6280 \).

05

Characterizing the Smallest 2% of Droplets

The smallest \( 2\% \) of droplets correspond to the 2nd percentile of the distribution. Find the \( z \)-score that labels the 2nd percentile, which is approximately \( -2.054 \). Use \( x = z \cdot \sigma + \mu \), substituting \( z = -2.054 \), \( \sigma = 150 \), and \( \mu = 1050 \): \( x = -2.054 \cdot 150 + 1050 = 739.9 \mu m \).

06

Probability of At Least One Drop Exceeding 1500 µm

For five independent droplets, calculate the probability that none exceed \( 1500 \mu m \), then subtract from 1 to find the probability of at least one exceeding. From Step 2, \( P(X < 1500) = 0.9987 \). Then, \( P(none > 1500) = 0.9987^5 = 0.9935 \). Thus, \( P(at\ least\ one > 1500) = 1 - 0.9935 = 0.0065 \).

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Key Concepts

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

Normal Distribution

A normal distribution is a bell-shaped curve that is symmetric around its mean. This is a common probability distribution used in statistics, characterized by its mean (average value) and standard deviation (measure of spread). It's used to represent real-valued random variables with a continuous probability distribution.
In the case of the droplet size measured in the exercise, the mean is given as \( 1050 \mu m \), and the standard deviation is \( 150 \mu m \). Together, these describe the normal distribution of droplet sizes, which allows for predictions of probabilities related to different sizes. Because many natural phenomena exhibit characteristics that can be modeled by this distribution, it's widely applicable.

Key Features of Normal Distribution:

• The total area under the curve is equal to 1, representing the total probability.
• About 68% of values fall within one standard deviation from the mean, 95% within two, and 99.7% within three standard deviations.
• It is defined by the parameters: mean (\( \mu \)) and standard deviation (\( \sigma \)).

Standard Normal Variable

The standard normal variable, often denoted as \( Z \), is a specific form of the normal distribution that has a mean of 0 and a standard deviation of 1. Converting other normal distributions to the standard form is beneficial because it allows us to use standard normal distribution tables to find probabilities.
To convert a normal distribution of any mean and standard deviation to the standard normal distribution, you use the z-score formula: \[ z = \frac{x - \mu}{\sigma} \]where \( x \) is the value from the normal distribution, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
For example, if you're calculating the probability that a droplet size is less than \( 1500 \mu m \), you'd compute \( z = \frac{1500 - 1050}{150} = 3 \). Using the z-score, you can then refer to a standard normal distribution table to find probabilities of different events.

Percentiles

Percentiles are measures used to indicate the value below which a given percentage of observations in a group fall. They are frequently used for comparing scores and values because they offer a relative standing within a dataset.
Being expressed as a percentage, they range from 0 to 100. For instance, finding the smallest 2% of droplets involves determining the 2nd percentile which corresponds to a particular z-score.
In this exercise, the z-score for the smallest 2% is approximately \( -2.054 \). To find the size of these droplets, this z-score can be transformed back into the original droplet size scale using the inverse of the z-score formula: \[ x = z \cdot \sigma + \mu \]Substituting the given variables, you'll find that the smallest 2% of droplets have a size of approximately \( 739.9 \mu m \). This process helps experts in determining threshold limits and making informed decisions.

Independent Events

Independent events are those whose outcomes do not affect each other. In probability, knowledge of the result of one event gives no information about the result of another; they are completely separate.
In the context of the exercise, the measurements of the sizes of droplets are independent events—each droplet size does not influence others.
To calculate the probability that at least one of five independently selected droplets exceeds \( 1500 \mu m \), you first find the probability of the opposite scenario (none exceeding \( 1500 \mu m \)) and subtract it from 1. This is done using:

• \( P(X < 1500) = 0.9987 \)
• The probability that none exceed is \( 0.9987^5 \)
• Finally, subtract this from 1 to find \( P(at\ least\ one > 1500) = 1 - 0.9935 = 0.0065 \)

Understanding independence helps predict outcomes in situations where multiple observations are made.