Question

A company makes a certain device. we are interested in the lifetime of the device. it is estimated that around 2% of the devices are defective from the start so they have a lifetime of 0 years. if a device is not defective, then the lifetime of the device is exponentially distributed with a parameter λ=2λ=2 years. let x be the lifetime of a randomly chosen device.

Answer 1

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Answer 2

Answer: The variance of a randomly chosen family is 0.495

Explanation:

$EX=\frac{1}{50}*0+\frac{49}{50}*EY$ where Y ~ Exponential(λ=2

$\frac{49}{50} *\frac{1}{2} =0.49$

$Var(X)=EX^{2} -(EX)^{2} =EX^{2} - (0.49)^{2}$

$EX=\frac{1}{50}*0+\frac{49}{50}*EY^{2} \\=\frac{49}{50}(\frac{1}{λ} +\frac{1}{λ^{2} } )$where $λ=2$

$=\frac{3}{4}* \frac{49}{50}$

Thus, $Var(X)=(\frac{3}{4}*\frac{49}{50})-(0.49)^{2} = 0.495$

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