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1 day ago

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Suppose a research paper states that the distribution of the daily sea-ice advance/retreat from each sensor is similar and is ap

proximately double exponential. The proposed double exponential distribution has density function f(x) = 0.5λe−λ|x| for −[infinity] < x < [infinity]. The standard deviation is given as 38.3 km. (Round your answers to four decimal places.) (a) What is the value of the parameter λ?

1 answer:

Svet_ta [4.3K]1 day ago

7 0

Answer:

The parameter λ has a value of 0.03692

Step-by-step explanation:

Take into account the given function.

$f(x) = 0.5\lambda e^{-\lambda |x|}$ for −∞ < x < ∞.

The standard deviation is stated to be 38.3 km.

Our task is to determine the parameter λ.

The formula for the probability density function of the double exponential distribution can be expressed as: $f(x)=\frac{e^{-|\frac{x-\mu}{\beta}|}}{2\beta}$

Here, μ represents the location parameter and β is the scale parameter.

When comparing the provided equation with the formula mentioned above, we have:

$\lambda=\frac{1}{\beta}$ and μ = 0.

The standard deviation = √2β

$S.D=\sqrt{2} \beta\\\beta=\frac{38.3}{\sqrt{2}}\\\beta=27.08219$

Next, insert the β value in $\lambda=\frac{1}{\beta}$ .

$\lambda=\frac{1}{27.08219}=0.03692$

Therefore, the value of parameter λ is 0.03692

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