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The paper "Microwave Obsevations of Daily Antarctic SeaIce Edge Expansion and Contribution Rates" (IEEE Geosci. and Remote Sensing Letters, 2006: 54-58) states that "The distribution of the daily sea-ice advance/retreat from each sensor is similar and is approximately double exponential." The proposed double exponential distribution has density function \(f(x)=.5 \lambda e^{-\lambda|x|}\) for \(-\infty

Short Answer

Expert verified
\(\lambda \approx 0.0346\); Probability within one standard deviation is approximately 0.865.

Step by step solution

01

Understanding the Distribution

The given function is a double exponential (Laplace) distribution with density function \(f(x) = 0.5 \lambda e^{-\lambda |x|}\). This distribution is symmetric and has its peak at the mean, which is 0 for a double exponential distribution around 0.
02

Standard Deviation of Double Exponential Distribution

The standard deviation for a double exponential distribution is \(\sigma = \frac{\sqrt{2}}{\lambda}\). Given that the standard deviation is 40.9 km, we have \(\frac{\sqrt{2}}{\lambda} = 40.9\).
03

Solving for \(\lambda\)

Using the equation from Step 2, \(\lambda = \frac{\sqrt{2}}{40.9}\). Calculate this to find the value of \(\lambda\).
04

Probability Calculation for \(|x|\leq \sigma\)

The probability that \(|x|\) is within one standard deviation is calculated as: \(P(-40.9 \leq x \leq 40.9) = 1 - 2(0.5)e^{-\lambda \cdot 40.9}\).
05

Calculating the Probability

Substitute the value of \(\lambda\) found in Step 3 into the probability equation from Step 4, and simplify to find the probability.

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Key Concepts

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

Probability Calculation
In statistics, calculating probability refers to determining how likely an event is to occur within a given distribution. For the double exponential distribution, we often need to find the probability that a variable lies within a certain range. This concept is essential when estimating outcomes and making predictions.

To compute the probability that a change in sea ice extent is within one standard deviation of the mean, we integrate the probability density function over the desired range. This means we're finding the area under the curve of the distribution from \(-40.9\) km to \(40.9\) km. It's common to express this probability as: \ P(-40.9 \leq x \leq 40.9) = 1 - 2(0.5)e^{-\lambda \cdot 40.9}\.

This formula accounts for the symmetry of the double exponential distribution, concentrating on how the distribution behaves around the mean, which is zero. Essentially, it sums the probabilities of a reduction or an expansion of the sea ice extent that lies within one standard deviation from the mean.
Parameter Estimation
Parameter estimation involves determining the parameters of a theoretical model or distribution that best fits the observed data. For the double exponential distribution, the key parameter is \(\lambda\), which influences the spread and height of the distribution curve. Estimating \(\lambda\) helps us understand the distribution's behavior, such as variability in the ice extent changes.

To estimate \(\lambda\) from the given standard deviation, we refer to the relationship \ \sigma = \frac{\sqrt{2}}{\lambda}\. Given the standard deviation \(\sigma\) is \(40.9\) km, solving \ \frac{\sqrt{2}}{\lambda} = 40.9\ results in \(\lambda = \frac{\sqrt{2}}{40.9}\). This value of \(\lambda\) then facilitates calculating probabilities and making predictions about sea ice extent distribution, as it quantifies the rate at which probability declines from the peak (mean).
Standard Deviation
Standard deviation is a measure of the spread of a set of values or, in probability distributions, the extent of deviation from the mean. For a double exponential distribution, the standard deviation is given by \(\sigma = \frac{\sqrt{2}}{\lambda}\). This relationship highlights the inverse relationship between \(\lambda\) and the spread of the distribution.

In our specific application involving sea ice extent, a standard deviation of \(40.9\) km indicates a relatively widespread variability around the mean, which is zero in this symmetric distribution. The higher the standard deviation, the more spread out the values are from the mean.

Understanding standard deviation is crucial when analyzing patterns, as it provides insight into the distribution's variability. It helps quantify how much individual data points can deviate from the mean, thus offering an understanding of stability or volatility in the context, such as daily changes in ice extent.
Probability Distribution
A probability distribution describes how the values of a random variable are distributed. For the double exponential distribution, also known as the Laplace distribution, it illustrates the probability of different outcomes related to a central peak. The density function \(f(x) = 0.5 \lambda e^{-\lambda |x|}\) represents this distribution.

This distribution is characterized by its peak at zero (in mean terms), and it exhibits a sharp central focus with exponential tails. In the context of sea ice extent, the double exponential distribution suggests that moderate changes occur frequently (high peak), while extreme changes are comparatively rare (thin tails).

Probability distributions like this one guide our understanding of data, enabling predictions and informing decisions. They provide a model that represents real-world phenomena, allowing scientists to interpret how changes in sea ice occur relative to their mean and within logical parameters of natural variability.

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Most popular questions from this chapter

Let \(X\) be the temperature in \({ }^{\circ} \mathrm{C}\) at which a certain chemical reaction takes place, and let \(Y\) be the temperature in \({ }^{\circ} \mathrm{F}\) (so \(Y=1.8 X+32\) ). a. If the median of the \(X\) distribution is \(\tilde{\mu}\), show that \(1.8 \tilde{\mu}+\) 32 is the median of the \(Y\) distribution. b. How is the 90th percentile of the \(Y\) distribution related to the 90 th percentile of the \(X\) distribution? Verify your conjecture. c. More generally, if \(Y=a X+b\), how is any particular percentile of the \(Y\) distribution related to the corresponding percentile of the \(X\) distribution?

If \(X\) is a normal rv with mean 80 and standard deviation 10 , compute the following probabilities by standardizing: a. \(P(X \leq 100)\) b. \(P(X \leq 80)\) c. \(P(65 \leq X \leq 100)\) d. \(P(70 \leq X)\) e. \(P(85 \leq X \leq 95)\) f. \(P(|X-80| \leq 10)\)

If \(X\) has an exponential distribution with parameter \(\lambda\), derive a general expression for the \((100 p)\) th percentile of the distribution. Then specialize to obtain the median.

a. If a normal distribution has \(\mu=30\) and \(\sigma=5\), what is the 91 st percentile of the distribution? b. What is the 6th percentile of the distribution? c. The width of a line etched on an integrated circuit chip is normally distributed with mean \(3.000 \mu \mathrm{m}\) and standard deviation \(.140\). What width value separates the widest \(10 \%\) of all such lines from the other \(90 \%\) ?

The automatic opening device of a military cargo parachute has been designed to open when the parachute is \(200 \mathrm{~m}\) above the ground. Suppose opening altitude actually has a normal distribution with mean value \(200 \mathrm{~m}\) and standard deviation \(30 \mathrm{~m}\). Equipment damage will occur if the parachute opens at an altitude of less than \(100 \mathrm{~m}\). What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

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