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Trees are subjected to different levels of carbon dioxide atmosphere with \(6 \%\) of them in a minimal growth condition at 350 parts per million (ppm), \(10 \%\) at 450 ppm (slow growth), \(47 \%\) at 550 ppm (moderate growth), and \(37 \%\) at 650 ppm (rapid growth). What are the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees in ppm?

Short Answer

Expert verified
Mean: 565 ppm, Standard Deviation: 82.92 ppm.

Step by step solution

01

Determine the Mean

To find the mean, multiply each condition by its probability and sum the results. Let the probabilities be \( p_1 = 0.06 \), \( p_2 = 0.10 \), \( p_3 = 0.47 \), \( p_4 = 0.37 \) and the conditions in ppm be \( x_1 = 350 \), \( x_2 = 450 \), \( x_3 = 550 \), and \( x_4 = 650 \). The mean \( \mu \) is given by:\[\mu = (p_1 \cdot x_1) + (p_2 \cdot x_2) + (p_3 \cdot x_3) + (p_4 \cdot x_4)\]Substitute the values:\[\mu = (0.06 \cdot 350) + (0.10 \cdot 450) + (0.47 \cdot 550) + (0.37 \cdot 650)\]Simplify:\[\mu = 21 + 45 + 258.5 + 240.5 = 565\]Thus, the mean is \( 565 \) ppm.
02

Calculate the Variance

The variance \( \sigma^2 \) is calculated using:\[\sigma^2 = p_1(x_1 - \mu)^2 + p_2(x_2 - \mu)^2 + p_3(x_3 - \mu)^2 + p_4(x_4 - \mu)^2\]Substituting the values:\[\sigma^2 = 0.06(350 - 565)^2 + 0.10(450 - 565)^2 + 0.47(550 - 565)^2 + 0.37(650 - 565)^2\]Calculate each term:\[0.06 imes 46225 = 2773.5\]\[0.10 imes 13225 = 1322.5\]\[0.47 imes 225 = 105.75\]\[0.37 imes 7225 = 2673.25\]Adding these results together:\[\sigma^2 = 2773.5 + 1322.5 + 105.75 + 2673.25 = 6875\]
03

Determine the Standard Deviation

The standard deviation \( \sigma \) is the square root of the variance:\[\sigma = \sqrt{6875} \approx 82.92\]Thus, the standard deviation is approximately \( 82.92 \) ppm.

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Key Concepts

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

Probability Distributions
Probability distributions are a way to model the likelihood of different outcomes of a random variable. In many real-world scenarios, like our exercise involving trees and carbon dioxide levels, it's critical to understand and calculate the probability of different growth conditions.

In this exercise, each growth condition, labeled by the carbon dioxide level in ppm, is associated with a probability, which represents the percentage of trees found in that condition. Here's a quick reminder of the probabilities used:
  • Minimal growth: 6% probability at 350 ppm
  • Slow growth: 10% probability at 450 ppm
  • Moderate growth: 47% probability at 550 ppm
  • Rapid growth: 37% probability at 650 ppm
To find the mean or expected carbon dioxide level experienced by the trees, we calculate the weighted average. This involves multiplying each possible outcome (in ppm) by its respective probability, then adding all these products together. This provides a central value that represents the probable condition of the environment across all trees.
Variance Calculation
Variance is a measure of how much the values of a random variable differ from the mean. In simple terms, it tells us how spread out the values are. The formula for variance used in our exercise is slightly adapted to account for probability, since not all conditions have the same likelihood. To calculate variance, follow these steps:
  • Subtract the mean from each condition,
  • Square the result for each condition,
  • Multiply each squared difference by its respective probability,
  • Sum all these values to obtain the variance.
Variance helps us quantify the variability in the carbon dioxide levels among different tree growth conditions. A high variance indicates that the conditions are widely spread around the mean (in high and low ppm), while a low variance suggests that they are more consistent, hovering closely around the mean.
Environmental Statistics
Environmental statistics involve using data analysis techniques to understand environmental data patterns and draw conclusions about the environment. By analyzing statistical properties such as the mean and standard deviation, we gain insights into environmental phenomena like tree growth in varied carbon dioxide levels. In our exercise, after calculating the mean and variance, we then determine the standard deviation, which is the square root of the variance. The standard deviation provides a more intuitive measure of spread as it is in the same units as the mean (ppm in this case). This measure tells us the typical distance of the carbon dioxide levels from the average, reflecting the consistency of growth conditions across the different atmospheric levels. Understanding these statistical parameters in environmental studies is vital for predicting and managing ecological conditions, assisting in making informed decisions about climate change impacts and environmental conservation efforts. By employing these tools, researchers can effectively monitor, interpret, and mitigate issues regarding atmospheric compositions and their effects on plant life.

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Most popular questions from this chapter

A player of a video game is confronted with a series of opponents and has an \(80 \%\) probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents. (a) What is the probability mass function of the number of opponents contested in a game? (b) What is the probability that a player defeats at least two opponents in a game? (c) What is the expected number of opponents contested in a game? (d) What is the probability that a player contests four or more opponents in a game? (e) What is the expected number of game plays until a player contests four or more opponents?

Traffic flow is traditionally modeled as a Poisson distribution. A traffic engineer monitors the traffic flowing through an intersection with an average of six cars per minute. To set the timing of a traffic signal, the following probabilities are used. (a) What is the probability that no cars pass through the intersection within 30 seconds? (b) What is the probability that three or more cars pass through the intersection within 30 seconds? (c) Calculate the minimum number of cars through the intersection so that the probability of this number or fewer cars in 30 seconds is at least \(90 \%\). (d) If the variance of the number of cars through the intersection per minute is \(20,\) is the Poisson distribution appropriate? Explain.

An air flight can carry 120 passengers. A passenger with a reserved seat arrives for the flight with probability 0.95. Assume that the passengers behave independently. (Use of computer software is expected.) (a) What is the minimum number of seats the airline should reserve for the probability of a full flight to be at least \(0.90 ?\) (b) What is the maximum number of seats the airline should reserve for the probability that more passengers arrive than the flight can seat to be less than \(0.10 ?\) (c) Discuss some reasonable policies the airline could use to reserve seats based on these probabilities.

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Suppose that \(X\) has a hypergeometric distribution with \(N=20, n=4,\) and \(K=4 .\) Determine the following: (a) \(P(X=1)\) (b) \(P(X=4)\) (c) \(P(X \leq 2)\) (d) Mean and variance of \(X\).

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