Question

CHAPTER 1,2/CLO 1,2

STAT102
14. Suppose a continuous random variable x has the probability density
0<x< 1
f(x) = {k(1- x),
10,
elsewhere
(a) Find k (b) Find P(0.1 < x < 0.2) (c) P(x > 0.5) using distribution function, determine
the probabilities that (d) x is less than 0.3 (e) between 0.4 and 0.6 (f) Calculate mean and
variance for the probability density function.​

Answer 1

a. In order for
`f` to be a proper density function, its integral over its domain must evaluate to 1:

`\displaystyle\int_(-\infty)^\infty f(x)\,\mathrm dx=k\int_0^1(1-x)\,\mathrm dx=\frac k2=1\implies k=2`

`f` also must be non-negative over its support, which is the case here.

I assume the instruction regarding the distribution function doesn't apply to parts (b) and (c).

(b) Integrate the density over the interval [0.1, 0.2]:

`P(0.1<x<0.2)=\displaystyle\int_(0.1)^(0.1)2(1-x)\,\mathrm dx=0.17`

(c) Integrate the density over the interval [0.5, 1]:

`P(X>0.5)=\displaystyle\int_(0.5)^12(1-x)\,\mathrm dx=0.25`

The distribution function is obtained by integrating the density:

`F(x)=\displaystyle\int_(-\infty)^xf(t)\,\mathrm dt=\begin{cases}0&\text{for }x<0\\2x-x^2&\text{for }0\le x<1\\1&\text{for }x\ge1\end{cases}`

(d) Using the distribution function, we have

`P(X<0.3)=F(0.3)=0.51`

(e) Using
`F` again, we get

`P(0.4<X<0.6)=P(X<0.6)-P(X<0.4)=F(0.6)-F(0.4)=0.84-0.64=0.2`

(f) The mean is

`E[X]=\displaystyle\int_(-\infty)^\infty x\,f(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13`

The variance is

`V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2`

where

`E[X^2]=\displaystyle\int_(-\infty)^\infty x^2\,f(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16`

so that the variance is

`V[X]=\frac16-\left(\frac13\right)^2=\frac1{18}`