Suppose x follows a distribution with density function: \begin{equation} f\left(x\right) = \left\{\begin{array}{rl} c\left|x - 2\right|,& 0 \le x \le 3\\ 0, & \text{otherwise}\\ \end{array}\right. \end{equation} (note: for this question you can enter your answer in decimals as well as fractions.) what must the value of c be so that f(x) is a probability density function? tries 0/5 find the cumulative distribution function of f(x) for $ 2 \leq x \leq 3 $. [ the accepted form of answer is an algebraic expression in terms of "x". all algebraically equivalent expressions to the correct answer are accepted. write product as *e.g 2*3 or 3*x, index/power as superscript,e.g 2^3 for 2 raised to the power 3, the exponential function as exp(x), the logarithm function as log(x) (and not as ln(x)) ] tries 0/5 find the median of the probability distribution of x tries 0/2 find e(x) tries 0/5 find the cumulative distribution function of f(x) for $ 0 \leq x \leq 2 $. [ the accepted form of answer is an algebraic expression in terms of "x". all algebraically equivalent expressions to the correct answer are accepted. write product as *e.g 2*3 or 3*x, index/power as superscript,e.g 2^3 for 2 raised to the power 3, the exponential function as exp(x), the logarithm function as log(x) (and not as ln(x)) ]

The question is a bit of a mess, so here's an attempt at parsing out the relevant information.

Given the PDF of a random variable [tex]X[/tex]:

[tex]f_X(x)=\begin{cases}c|x-2|&\text{for }0\le x\le3\\0&\text{otherwise}\end{cases}[/tex]

Find [tex]c[/tex], the CDF, the median of [tex]X[/tex], and the expectation of [tex]X[/tex] (in no particular order).

For [tex]f_X(x)[/tex] to be valid PDF, we require

[tex]\displaystyle\int_{-\infty}^\infty f_X(x)\,\mathrm dx=1[/tex]

Note that

[tex]|x-2|=\begin{cases}x-2&\text{for }x\ge2\\2-x&\text{for }x<2\end{cases}[/tex]

We have

[tex]\displaystyle\int_{-\infty}^\infty f_X(x)\,\mathrm dx=c\int_0^3|x-2|\,\mathrm dx=c\int_0^2(x-2)\,\mathrm dx+c\int_2^3(2-x)\,\mathrm dx[/tex]
[tex]\implies\dfrac{5c}2=1\implies c=\dfrac25[/tex]

The CDF is defined by

[tex]F_X(x)=\mathbb P(X\le x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt[/tex]

To find the CDF, we compute the integral above for the four possible cases:

[tex]x<0\implies\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=0[/tex]
[tex]0\le x<2\implies\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=c\int_0^x|t-2|\,\mathrm dt=\frac{4x-x^2}5[/tex]
[tex]2\le x<3\implies\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\frac45+c\int_2^x|t-2|\,\mathrm dt=\frac45-c\int_2^x(2-t)\,\mathrm dt=\frac{8-4x+x^2}5[/tex]
[tex]x\ge3\implies\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=1[/tex]

So the CDF is

[tex]F_X(x)=\begin{cases}0&\text{for }x<0\\\\\dfrac{4x-x^2}5&\text{for }0\le x<2\\\\\dfrac{8-4x+x^2}5&\text{for }2\le x<3\\\\1&\text{for }x\ge3\end{cases}[/tex]

The median is the value [tex]M[/tex] such that [tex]\mathbb P(X\le M)=\dfrac12[/tex]. From the PDF, we can gather that the median must fall somewhere in [tex]0\le x\le2[/tex]. The CDF then tells us that

[tex]\mathbb P(X\le M)=F_X(M)=\dfrac12=\dfrac{4M-M^2}5\implies M=2-\sqrt{\dfrac32}\approx0.775[/tex]

The expected value of [tex]X[/tex] is given by

[tex]\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx[/tex]

We have

[tex]\mathbb E[X]=\displaystyle\frac25\int_0^2x(x-2)\,\mathrm dx+\frac25\int_2^3x(2-x)\,\mathrm dx=\frac{16}{15}[/tex]