How Does Exponential Decay Model the Lifespan of Computer Chips?

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The discussion revolves around the application of exponential decay to model the lifespan of computer chips, using the probability function 0.00005∫e^(-0.00005t)dt. Participants seek to calculate the probability of chip failure after 15,000 hours and the fraction of chips that continue to operate for at least another 15,000 hours after that point. There is emphasis on correctly interpreting the probability statements and the need for conditional probability in the second query. Additionally, the importance of simplifying notation and performing the integral before answering specific questions is highlighted. Overall, the conversation aims to clarify the mathematical approach to understanding chip longevity through exponential decay.
Eleni
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Homework Statement


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.

Homework Equations


0.00005∫e^(-0.00005t)dt [a, ∝]
a= hours of operation

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.
 
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Assuming that by ∝ you mean ∞:
Eleni said:
i. do I evaluate P(0≤ x≤15000)= 0.00005∫e^(-0.00005t)dt on the interval [0, 15000]
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.)
Eleni said:
ii. do I evaluate P(15000≤ x≤ 30000)= 0.00005∫e^(-0.00005t)dt on the interval [15000, 30000]
##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.
Eleni said:
iii. would the meaning of this integral evaluation be the total operating life expectancy of the computer chip (hrs)?
No, you compute the probability that the chip will fail before ...
 
Last edited:
Eleni said:

Homework Statement


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.

Homework Equations


0.00005∫e^(-0.00005t)dt [a, ∝]
a= hours of operation

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.

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

You have that the time to failure, ##X##, satisfies
P(X > a) = \int_a^{\infty} r e^{-rt} \, dt
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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