Question :
The length of time(in hundreds
The length of time(in hundreds of hours) for the failure of a transistor is a random variable Y with distribution function ;
F(Y) = { 0 : if y<0 or [1 - e^(-(y^2) ] if y>=0 }
Find the probability that the transistor operates for at least 200 hours.
Since we need to find the probability that the transistor operates for at least 200 hours, we nedd to find, P(Y>=200) = integrate [f(t) dt] from -infinity to 200.
P(Y>=200) = integrate [ 0 dt] from -infinity to 0 + integrate [1 - (e^(-(t^2))) dt] from 0 to 200
==> P(Y>=200) = 0 + integrate [1 - (e^(-(t^2))) dt] from 0 to 200
==> P(Y>=200) = integrate [1 - (e^(-(t^2))) dt] from 0 to 200
==> P(Y>=200) = integrate [1 dt ] from 0 to 200 + integrate [ - (e^(-(t^2))) dt] from 0 to 200
But, how do we the following integration? I'm having trouble finding a suitable substitution or any other method for the following integration ; I = integrate [1 - (e^(-(t^2))) dt] from 0 to 200
Or is there any other method we can use to do this problem?
Thank you!
To solve the integration problem, you can try using a technique called integration by parts. Here's how you can do it:
Let I = ∫[1 - e^(-t^2)] dt from 0 to 200.
Integration by parts is given by the formula:
∫[u dv] = uv - ∫[v du]
Let's choose u = t and dv = -e^(-t^2) dt.
Therefore, du = dt and v = ∫[-e^(-t^2)] dt.
To find v, you can make the substitution w = -t^2, so dw = -2t dt:
∫[-e^(-t^2)] dt = ∫e^w (-1/2)dw = -(1/2)∫e^wdw.
Now we can integrate to get v:
v = -(1/2)∫e^wdw = -(1/2)e^w = -(1/2)e^(-t^2).
Applying the integration by parts formula, we have:
I = uv - ∫[v du]
I = t * (-(1/2)e^(-t^2)) - ∫[-(1/2)e^(-t^2) dt]
I = -(1/2)te^(-t^2) + (1/2)∫e^(-t^2) dt.
This new integral is a special function called the error function, denoted as erfi(t). Therefore, the integral simplifies to:
I = -(1/2)te^(-t^2) + (1/2)erfi(t).
Now we can evaluate the integral over the range [0, 200]:
P(Y >= 200) = I evaluated from 0 to 200
P(Y >= 200) = [-(1/2)(200)e^(-200^2) + (1/2)erfi(200)] - [-(1/2)(0)e^(-0^2) + (1/2)erfi(0)].
Since e^(-0^2) = e^0 = 1 and erfi(0) = 0, this simplifies further to:
P(Y >= 200) = [-(1/2)(200)e^(-200^2) + (1/2)erfi(200)] - 0.
Finally, you can compute the approximate numerical value using a calculator or computer software that can compute the error function.