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Integration application

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data
    The probability that a particular computer chip fails after a hours of operation is given by
    0.00005∫e^(-0.00005t)dt on the interval [a, ∝]
    i. find the probability that the computer chip fails after 15,000 hours of operation
    ii. of the chips that are still operating after 15,000 hours, what fraction of these will operate for at least another 15,000 hours?
    iii. Evaluate 0.00005∫e^(-0.00005t)dt [0, ∝]t and interpret its meaning in the context of the situation.
    2. Relevant equations
    0.00005∫e^(-0.00005t)dt [a, ∝]
    a= hours of operation
    3. The attempt at a solution
    i. do I evaluate P(0≤ x≤15000)= 0.00005∫e^(-0.00005t)dt on the interval [0, 15000]
    ii. do I evaluate P(15000≤ x≤ 30000)= 0.00005∫e^(-0.00005t)dt on the interval [15000, 30000]
    iii. would the meaning of this integral evaluation be the total operating life expectancy of the computer chip (hrs)?

    please accept my apologies if this appears quite messy. I am not very good at using latex or typing equations and would much rather the old school pen and paper.
    Please note: -0.00005t is the supercript for e in the formula

    THANK YOU.
     
  2. jcsd
  3. Dec 29, 2015 #2

    Samy_A

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    Assuming that by ∝ you mean ∞:
    Yes, but why? What is the formula for ##P(X \leq 15000)##?
    (Here I interpret "fails after 15,000 hours of operation" as you did, meaning "fails after at most 15,000 hours of operation". It could also mean "fails after at least 15,000 hours of operation", though.)
    ##P(15000≤ X≤ 30000)## is the probability that a chip fails between 15000 and 30000 hours. That is not what you is asked for.
    You need to evaluate ##P(X \geq 30000|X \geq15000)##, a conditional probability.
    No, you compute the probability that the chip will fail before ...
     
    Last edited: Dec 29, 2015
  4. Dec 29, 2015 #3

    Ray Vickson

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    First hint (for problem solving in general): simplify notation!

    You have that the time to failure, ##X##, satisfies
    [tex] P(X > a) = \int_a^{\infty} r e^{-rt} \, dt [/tex]
    where ##r = 0.00005 = .5 \times 10^{-4} = 1/20000## (##\text{hr}^{-1}##). Even if you don't know LaTeX (or care to learn it), the expression "int{e^(-rt) dt, t=a..infinity}" is still a lot easier to type than what you gave, and is easier to read as well. Isn't it a lot better to type "r" than "0.00005"? Anyway, if ##t## is in 10s of thousands of hours, writing 0.00005t as t/20000 is more "revealing": when t = 15000 the product is rt = 15/20 = 0.75, and when t = 30000 it is rt = 30/20 = 1.5.

    Second hint: do the integral first, before doing anything else.

    As for your questions: (i) and (ii) what do YOU think, and why? Can you write the answer to (ii) in terms of ##P(X > 30000)## and ##P(C > 15000)##? For question (iii): do not try to answer it until you have actually done the integral, as I suggested.

    Note that PF rules forbid us from being more helpful until you have done some work on the problem: give us your answers first, then we can talk.
     
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