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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 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 exactly two of them exceed \(1500 \mu \mathrm{m}\) ?

Step by step solution

01

Standardize the droplet size 1500 μm

To find the probability that a droplet is less than 1500 μm, we first standardize this value using the z-score formula. The z-score is calculated by \[ z = \frac{x - \mu}{\sigma} \] where \(x \) is the droplet size (1500 μm), \(\mu \) is the mean (1050 μm), and \(\sigma \) is the standard deviation (150 μm). Thus, \[ z = \frac{1500 - 1050}{150} = \frac{450}{150} = 3 \].

02

Find the probability for z-score of 3

Using a z-table, or normal distribution calculator, find the probability corresponding to a z-score of 3. From the z-table, the probability \( P(Z < 3) \) is approximately 0.9987. Therefore, the probability that the size of a single droplet is less than 1500 μm is 0.9987.

03

Standardize droplet size 1000 μm

Next, calculate the z-score for a droplet size of 1000 μm using the same formula. \[ z = \frac{1000 - 1050}{150} = \frac{-50}{150} = -0.3333 \].

04

Find the probability for z-score of -0.3333

Using the z-table, find the probability for \(z\) being greater than -0.3333 (i.e., \(P(Z > -0.3333)\)). This can be found using the fact that \(P(Z > z) = 1 - P(Z < z)\). From the z-table, \(P(Z < -0.3333) \approx 0.3707\). Hence, \(P(Z > -0.3333) = 1 - 0.3707 = 0.6293\). The probability that the droplet size is at least 1000 μm is 0.6293.

05

Calculate probability for droplet size between 1000 and 1500 μm

Calculate the probability for a droplet size between 1000 and 1500 μm by taking the difference:\[ P(1000 < X < 1500) = P(Z < 3) - P(Z < -0.3333) = 0.9987 - 0.3707 = 0.6280 \].

06

Characterize the smallest 2% of droplets

To find the droplet size that corresponds to the smallest 2%, we need the z-score that corresponds to the 2nd percentile, which is approximately -2.05 from the z-table. Use the z-score formula to compute the size.\[ -2.05 = \frac{x - 1050}{150} \].Solving for \(x\), we get:\[ x = -2.05 \times 150 + 1050 = 741.5 \]. Thus, the smallest 2% of droplet sizes are those less than 741.5 μm.

07

Probability of exactly two droplets exceeding 1500 μm

First find the probability that the size of a droplet exceeds 1500 μm, which is \[ P(X > 1500) = 1 - P(X < 1500) = 1 - 0.9987 = 0.0013 \]. Use the binomial probability formula:\[ P(X = 2) = \binom{5}{2} \cdot (0.0013)^2 \cdot (1 - 0.0013)^3 \].Calculating:\[ \binom{5}{2} = 10 \quad (0.0013)^2 = 0.00000169 \quad (0.9987)^3 \approx 0.9961 \].Therefore:\[ P(X = 2) = 10 \times 0.00000169 \times 0.9961 \approx 0.0000168 \].

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Key Concepts

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

Understanding the Z-Score

The z-score is a fundamental concept when dealing with the normal distribution. It helps us understand how far a particular value is from the mean, in terms of the number of standard deviations. To calculate the z-score, use the formula \( z = \frac{x - \mu}{\sigma} \), where \( x \) is the value you are interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For instance, in the context of spray droplet sizes, if we want to find out how unusual a droplet size of 1500 μm is, knowing the mean size is 1050 μm and the standard deviation is 150 μm, the calculation would be \( z = \frac{1500 - 1050}{150} = 3 \).
This z-score of 3 indicates that a 1500 μm droplet is three standard deviations above the mean size. The usefulness of the z-score lies in its ability to help us utilize standard normal distribution tables or calculators to find probabilities for our normal data.

• The formula standardizes different data by providing a common point of reference.
• It allows easy look-up of probabilities in standard normal tables.

Probability Calculation in Normal Distribution

Calculating probability for a normal distribution is about figuring out how likely it is that a value or range of values will occur. This is where z-scores and z-tables come into play. A normal distribution is a bell-shaped curve that is symmetrical around the mean.
Once you have the z-scores for the data points of interest, you can reference a z-table. For example, with a z-score of 3 (as calculated for a 1500 μm droplet), the probability of a size less than 1500 μm can be found. A z-score of 3 corresponds to a very high probability, around 0.9987.
To find the probability of a droplet being between two sizes, say between 1000 μm and 1500 μm, you subtract the smaller z-value probability from the larger one. With z-scores of -0.3333 and 3 respectively, the probability ends up being 0.9987 - 0.3707 = 0.6280.

• Z-tables convert z-scores into probabilities.
• Subtraction of probabilities gives probability for an interval.

The Concept of Binomial Probability

Binomial probability is useful when you want to find the likelihood of a certain number of successes in a given number of independent trials, where each trial has two possible outcomes: success or failure. The formula is \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( n \) is the number of trials, \( k \) is the number of successes, and \( p \) is the probability of success on a single trial.
Relating to droplet sizes exceeding 1500 μm, if we're checking five droplets, the probability of any single droplet exceeding that is 0.0013. To find the probability that exactly two of these exceed 1500 μm, you plug into the binomial formula: \( P(X = 2) = \binom{5}{2} \times (0.0013)^2 \times (0.9987)^3 \). With calculations, this results in a tiny probability of approximately 0.0000168.

• Good for discrete events with binary outcomes.
• Applies fundamental counting principle with probability.