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An article in a magazine discussed the length of time till failure of a particular product. At the end of theâ product's lifetime, the time till failure is modeled using an exponential distribution with a mean of thousand hours. In reliability jargon this is known as theâ "wear-out" distribution for the product. During its normalâ (useful) life, assume theâ product's time till failure is uniformly distributed over the range thousand to million hours. Complete parts a through c.

a. At the end of the productâs lifetime, find the probability that the product fails before 700 thousand hours.
b. During its normal (useful) life, find the probability that the product fails before 700 thousand hours.
c. Show that the probability of the product failing before 830 thousand hours is approximately the same for both the normal (useful) life distribution and the wear-out distribution.

Sagot :

ajeigbeibraheem

Answer:

Step-by-step explanation:

From the missing information;

Lets assume that;

[tex]\text{the exponential distribution mean = 500 thousand naira}[/tex]

[tex]\text{Range 100 thousand}[/tex] → [tex]\text{1 million naira}[/tex]

∴

a)

[tex]X \sim exp (\dfrac{1}{500}) \\ \\ P( X \le x) = 1 - e^{\dfrac{-x}{500}} \\ \\ P(X < 700) = 1 - e^{\frac{-7}{5}} \\ \\ \mathbf{P(X< 700) = 0.75340}[/tex]

b)

[tex]Y \sim \cup (100,1000) \implies P(Y \le y ) = \dfrac{y - 100}{1000-100} \\ \\ P(Y < 700) = \dfrac{600}{900} = 0.67[/tex]

c)

[tex]P(X < 830)= 1 - e^{\frac{-830}{500}} \\ \\ P(X < 830)= 1 - 0.1901 \\ \\ P(X < 830)=0.8099 \\ \\ P(Y < 830) = \dfrac{730}{900} = 0.8111 \\ \\ Thus;\mathbf{ P(X < 0.8099) = P(Y< 0.8111)}[/tex]

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