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Question
Solve the following problem :
Determine k if the p.d.f. of the r.v. is
f(x) = `{("ke"^(-thetax), "for" 0 ≤ x < oo),(0, "otherwise".):}`
Find `"P"("X" > 1/theta)` and determine M is P(0 < X < M) = `(1)/(2)`
Solution
Since f(x) is the p.d.f. of X.
∴ `int_0^oo f(x)*dx` = 1
∴ `int_0^oo ("ke"^(-thetax))*dx` = 1
∴ `"k"*[("e"^(-thetax))/(-theta)]_0^oo` = 1
∴ `-"k"/theta ["e"^(-thetax))]_0^oo` = 1
∴ `- "K"/theta [1/"e"^(thetax)]_0^oo` = 1
∴ `-"k"/theta(1/oo - 1/1)` = 1
∴ `-"k"/theta(0 - 1)` = 1
∴ k = `theta`
F(x) = `"k" int_0^x "e"^(-thetax)*dx`
= `theta int_0^x e^(-thetax)*dx` ...[∵ k = θ]
= `theta[("e"^(-thetax))/-theta]_0^x`
= `-["e"^(-thetax)]_0^x`
= `- ("e"^(-thetax) - 1)`
= `1 - (1)/"e"^(thetax)` ...(i)
∴ `"P"("X" > 1/theta) = 1 - "P"("X" ≤ 1/theta)`
= `1 - "F"(1/theta)`
= `1 - [1 - 1/"e"^(theta(1/theta))]` ...[From (i)]
= `(1)/"e"`
Given that, P(0 < X < M) = `(1)/(2)`
∴ F(M) – F(0) = `(1)/(2)`
∴ `1 - 1/"e"^(theta"M") - 0 = (1)/(2)` ...[From (i)]
∴ `(1)/(2) = (1)/"e"^(theta"M")`
∴ eθM = 2
∴ θM = log 2
∴ M = `(1)/theta log 2`.
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