Independent solution

How to solve this Mixture Distributions question

Setup

Setup

Integrate the positive-support density to obtain its cumulative distribution function. The two exponential terms carry different integration factors.

F(x)=∫0x0.4(e−0.6t+e−1.2t) dtF(x)=\int_0^x 0.4\left(e^{-0.6t}+e^{-1.2t}\right)\,dt
F(x)=1−23e−0.6x−13e−1.2x,x>0F(x)=1-\frac{2}{3}e^{-0.6x}-\frac{1}{3}e^{-1.2x},\qquad x>0

Model

Model

At a median m, the CDF and survival probability are both one-half. Substituting y equal to the slower exponential term converts the equation to a quadratic.

23e−0.6m+13e−1.2m=12\frac{2}{3}e^{-0.6m}+\frac{1}{3}e^{-1.2m}=\frac{1}{2}
y=e−0.6m⟹23y+13y2=12y=e^{-0.6m}\quad\Longrightarrow\quad \frac{2}{3}y+\frac{1}{3}y^2=\frac{1}{2}

Compute

Compute

Retain the positive quadratic root because an exponential value must lie between zero and one, then recover m with a logarithm.

2y2+4y−3=0⟹y=10−222y^2+4y-3=0\quad\Longrightarrow\quad y=\frac{\sqrt{10}-2}{2}
m=−10.6log⁡ ⁣(10−22)=0.9046093341…m=-\frac{1}{0.6}\log\!\left(\frac{\sqrt{10}-2}{2}\right)=0.9046093341\ldots

Answer

Answer

The median rounds to 0.905 at the precision used by the answer list.

m≈0.905(B)\boxed{m\approx 0.905\quad\text{(B)}}