The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine the probability that

a. A component lasts more than 3000 hours before failure.
b. A component fails in the interval from 1000 to 2000 hours.
c. A component fails before 1000 hours.
d. Determine the number of hours at which 10% of all components have failed.

Step-by-step explanation: Probability Density Function is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = [tex]\int\limits^b_a {P(x)} \, dx[/tex]

Then, for the electronic component, probability will be:

P(a<x<b) = [tex]\int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx[/tex]

P(a<x<b) = [tex]\frac{1000}{1000}.e^{\frac{-x}{1000} }[/tex]

P(a<x<b) = [tex]e^{\frac{-b}{1000} }-e^\frac{-a}{1000}[/tex]

a. For a component to last more than 3000 hours:

P(3000<x<∞) = [tex]e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}[/tex]

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = [tex]e^{-3}[/tex]

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = [tex]e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}[/tex]

P(1000<x<2000) = [tex]e^{-2}-e^{-1}[/tex]

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = [tex]e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}[/tex]

P(0<x<1000) = [tex]e^{-1}-1[/tex]

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = [tex]e^{\frac{-b}{1000} }-e^\frac{-a}{1000}[/tex]

0.1 = [tex]1-e^\frac{-x}{1000}[/tex]

[tex]-e^{\frac{-x}{1000} }=-0.9[/tex]

[tex]{\frac{-x}{1000} }=ln0.9[/tex]

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.