The probability density function of X, the lifetime of a certain type of electronic device (in hours), is given byf(x)={10x2x&gt;100x≤10&nbsp;(a) Find&nbsp;{X&gt;20}.&nbsp;(b) What is the cumulative distribution function of X?&nbsp;(c) What is the probability that, of 6 such types of devices, at least 3 will function for at least 15 hours? What assumptions are you making?

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The probability density function of X, the lifetime of a certain type of electronic device (in hours), is given byf(x)={10x2x&amp;gt;100x≤10&amp;nbsp;(a) Find&amp;nbsp;{X&amp;gt;20}.&amp;nbsp;(b) What is the cumulative distribution function of X?&amp;nbsp;(c) What is the probability that, of 6 such types of devices, at least 3 will function for at least 15 hours? What assumptions are you making?

Donald Proulx · Accepted Answer

P(X&amp;gt;20)=1−∫102010x2&amp;nbsp;dx&amp;nbsp;=1−10∫10201x2&amp;nbsp;dx&amp;nbsp;=1+10⋅1x∣1020=1+10⋅(120−110)=1−10⋅120=12Cummulative distribution function of X is&amp;nbsp;Fx(x)=P(X≤x):FX(x)=P(X≤x)=∫10x10a2da=−10⋅1a∣10x=−10⋅(1x−110)=1−10x,where&amp;nbsp;x&amp;gt;10&amp;nbsp;by the assumption of the assigment. If the lifetime of some electronic device is f, we want to find the probability that out of 6 such devices, 3 will function at least 15 hours. Let A denonte the number of the devices that work:𝟚𝟙P(A≥3)=1−A≤2=1−P(A=0)−A=1−P(A=2)P(A=0)=(60)⋅P(X&amp;lt;15)6=136P(A=1)=(61)⋅23⋅135

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The probability density function of X, the lifetime of a certain type of electro

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