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Consider the probability density function \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\) and \(\lambda > 0\). Determine the value of \(\lambda\) such that \(5E(x) = V(x)\).
in Probability recategorized by
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Exponential Distribution

f(x) = $\lambda$e$^{-\lambda x}$

Mean in exponential distribution = 1/$\lambda$ = E(X)

Variance = 1/$\lambda ^{2}$

5/$\lambda$ = 1/$\lambda ^{2}$ 

$\lambda$ = 1/5 or 0.2

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