Calculating Probability for Exponential Distribution in Unplanned Shutdowns

AI Thread Summary
The discussion revolves around calculating the probability of time between unplanned shutdowns of a power plant, modeled by an exponential distribution with a mean of 20 days. The initial confusion stems from determining the correct value for X in the equation, leading to the realization that the probability of time being less than 21 days is given by 1 - e^(-20)(21). After some calculations, the participant arrives at a final probability of 0.150, suggesting that the time between shutdowns is likely to exceed 21 days. The participant seeks validation of their calculations against provided answer choices, expressing confidence in their final answer. The conversation highlights the importance of understanding exponential distribution in practical applications.
Changoo
Messages
7
Reaction score
0
I am having a lot of trouble with a homework question from my book. It asks:

The time between unplanned shutdowns of a power plant has an exponential distribution with a mean of 20 days. Find the probability that the time between two unplanned shutdowns is more than 21 days.

I know this much so far 1-e-(20)(?) (one minus e to the negative power of mean times any value of the continuous variable(X))

I am lost on finding X within the equation.

Hope someone can help.
 
Physics news on Phys.org
Hello!

First of all what do we define as an exponential distribution? It is equaled to:

1 - e^(-yx)

= 1 - e^(-20)(21)

Since this is the probability that the time between two unplanned shutdowns is less than 21 we don't need the 1. Hence our answer (I think) would be:

e^(-20)(21) or essentially 0.

Hope this helps!
 
Last edited:
Thanks for your help,

But how do I solve E^-(20)(21)? I know that 20 times 21 is 420. How do I determine e^-420?

The book has given me some answers, but none say zero. a. .350, b. .650, c. .150, d. .850

I can probabily figure out the answer with no problem if someone can help me witht the problem above.
Thanks for your help!
 
Okay, here is what I have, please tell me if I am right:

F(x)=1-e^-(20)(2/21)

F(x)=1-.850

F(x)=.150 (final answer)

I hope I am write.
 
I meant right**** Sorry :blushing:
 
Asking for Review

I feel confident about my answer, I am hoping someone can review and let me know if I have calculated wrong in any way. :approve:
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top