Probability that a bulb lasts for at least 7 months

[jaus tail]

Homework Statement

Homework Equations

I don't even know how to start this. Should I use Poisson's distribution. Or assume f(x) = exp(-x)
And then since mean is given we have integration from 0 to infinite xf(x) is mean.
Not sure what this will give.

The Attempt at a Solution

Let f(x) be probability that bulb won't fuse in month x.
So f(x) = ke^-x where x is months. k is some constant
So integrating from 0 to infinity f(x) should be 1 as all probability integration/summation is 1
This gives k = 1
So f(x) = e^-x
Now integrating this from 0 to 6 I get: 0.99.
I subtract this from 1. I get 0.002478. This doesn't match book answer which is 0.2466

 

[mjc123]

No, you don't use Poisson's distribution; you use the exponential distribution, because you are told it's exponential.
The probability of the lifetime being between t and t+dt (i.e. the probability that it fails during this interval) is f(t)dt, where
f(t) = (1/τ)*e^-t/τ where τ is a constant
What is the mean lifetime of this distribution? (If you don't know this for the exponential distribution, it is easy to work out).
What is the probability of the lifetime being longer than a given value of t?

 

[jaus tail]

Why is constant same for exponential power and it's coefficient?
Mean is integrating xf(x) from limits 0 to infinity.
Taking constant k, I get to find mean:
Integrating: ke^(-kt)dt from limits 0 to infinity. This equals 5--> Given in question.
So I do this and get value of k as 1/5
f(life of bulb) = f(bulb fuses in time x) is 1/5 e^(-5t)
Now to find probability that bulb lasts for atleast 7 months. This is 1 - P(bulb fuses in this time) = 1 - integrating f(x) from limits 0 to 7
∫(1/5)e^-5tdt limits 0 to 7
which is 0.7534

So 1 - 0.7534 gives 0.2466
Matches book answer.
But why did you take constant in exponential part same as constant in coefficient part?

 

[mjc123]

Because that's how it works out. If you let f(t) = Ae^-kt and integrate it from 0 to infinity, the integral equals 1 if A = k.

 

[Ray Vickson]

Why is constant same for exponential power and it's coefficient?
Mean is integrating xf(x) from limits 0 to infinity.
Taking constant k, I get to find mean:
Integrating: ke^(-kt)dt from limits 0 to infinity. This equals 5--> Given in question.
So I do this and get value of k as 1/5
f(life of bulb) = f(bulb fuses in time x) is 1/5 e^(-5t)
Now to find probability that bulb lasts for atleast 7 months. This is 1 - P(bulb fuses in this time) = 1 - integrating f(x) from limits 0 to 7
∫(1/5)e^-5tdt limits 0 to 7
which is 0.7534

So 1 - 0.7534 gives 0.2466
Matches book answer.
But why did you take constant in exponential part same as constant in coefficient part?

Why did YOU write $k e^{-kt}?$ BTW: that should be $(1/5) e^{-t/5}$, not $(1/5) e^{-5t}$ as you wrote.

 

[Mark44]

Thread moved. Any question that involves integration or differentiation should be posted in the Calculus & Beyond section.

 

[StoneTemplePython]

No, you don't use Poisson's distribution; you use the exponential distribution, because you are told it's exponential.

This statement is extremely narrow to the point I'd call it wrong.
- - - - - - - - - - - - - - -

A Poisson process is a counting process with exponential interrarival times. The problem can be solved with the use

(a) integrating the exponential distribution PDF, which is what was done and the least efficient method
(b) using the (complement of) CDF of the exponential distribution, with $\lambda = \frac{1}{5}$ and $\tau = 7$
(c) using a Poisson distribution with $\lambda = \frac{1}{5}$, $\text{time} =\tau = 7$ and $k_{arrivals} = 0$

in general we can see that the complement of the CDF of the exponential distribution in (b) is equivalent to a counting process of exponentials with zero arrivals at time $\tau$ i.e. (c).

I.e. for that one arrival with some fixed $\lambda$, with arrival time $T$

$P(T \gt \tau) = 1 - CDF_{\lambda}(\tau) = P_{\lambda}(0, \tau\big)$

If OP is able to draw the picture and understand what it means for a counting process to have had zero arrivals (i.e. not even one light bulb burnt out) at time $\tau$ and each (inter)arrival is exponentially distributed, then yes using a Poisson is in fact how I'd suggest this problem be answered.

No integration needed.

 

[jaus tail]

Thanks. I hadn't taken the k part inside the exponential curve earlier. I think it's same if you write k or 1/(constant). In first case I got k as 1/5, in second case I'll get 5.