The pdf describing the chance that a given insect will die at a particular age, x days, is f(x)=4.1x10^-2 xe^-.2x. What is the chance the insect will live between 2 and 4 days?
To find the probability that the insect will live between 2 and 4 days, we need to calculate the integral of the probability density function (PDF) between these two limits.
The PDF given is f(x) = 4.1x10^-2 xe^(-0.2x).
To find the probability, we integrate the PDF over the range from 2 to 4:
P(2 ≤ x ≤ 4) = ∫[2,4] f(x) dx
Let's calculate this integral step by step:
First, multiply the terms inside the PDF:
f(x) = 4.1x10^-2 xe^(-0.2x) = 0.041x e^(-0.2x)
Now, integrate the function with respect to x from 2 to 4:
P(2 ≤ x ≤ 4) = ∫[2,4] 0.041x e^(-0.2x) dx
To solve this integral, we can use integration by parts. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
Let's assign u = x and dv = 0.041e^(-0.2x) dx. Then, du = dx and v = ∫ 0.041e^(-0.2x) dx.
The integral v can be solved separately:
v = ∫ 0.041e^(-0.2x) dx
To integrate this, we can use substitution. Let z = -0.2x, so dz = -0.2 dx.
v = ∫ e^z dz = e^z = e^(-0.2x)
Now, let's apply integration by parts:
∫ 0.041xe^(-0.2x) dx = 0.041x e^(-0.2x) - ∫ e^(-0.2x) dx
The remaining integral can be easily solved:
∫ e^(-0.2x) dx = (-1/0.2) e^(-0.2x) = -5e^(-0.2x)
Now, substituting the bounds [2,4], we have:
P(2 ≤ x ≤ 4) = [0.041x e^(-0.2x) -(-5e^(-0.2x))] evaluated from 2 to 4
P(2 ≤ x ≤ 4) = [0.041(4) e^(-0.2(4)) -(-5e^(-0.2(4)))] - [0.041(2) e^(-0.2(2)) -(-5e^(-0.2(2)))]
Calculating this expression will give us the probability that the insect will live between 2 and 4 days.