Question

(dispersion) DA), standard error σ(X) , asymmn etry coefficient ASA) an 2. Calculate multiplier k. Find distribution function fe), mode Mok), median Mea), expectation value M(x), variance (dispersion) D(x), standard error σ(x), asymmetry coefficient As(x) and excess Ex(x) for continuous distributions with the given probability densities a) b) x>0 x-16sxs0 x20sxs-16 x<-20 x <2 160 f(x)=1 0 r>6 40 2 Calculate probability that X E-16:6] Calculate probability that X E [1:4]

Answer 1

b) Consider the PDF $\rho \left ( x \right )=k\left ( 8x-x^2-12 \right );2\leqslant x\leqslant 6$ .

The condition for PDF is

$\int_{2}^{6}\rho \left ( x \right )dx=1\\ \int_{2}^{6}k\left ( 8x-x^2-12 \right )dx=1\\ k\left ( 4x^2-x^3/3-12x \right ]_{2}^{6}=1\\ {\color{Blue} k=3/32}$

The distribution function is

$F\left ( x \right )=\int_{2}^{x}\rho \left ( x \right )dx\\ F\left ( x \right )=\int_{2}^{x}k\left ( 8x-x^2-12 \right )dx\\ F\left ( x \right )=k\left ( 4x^2-x^3/3-12x \right ]_{2}^{x}\\ {\color{Blue} F\left ( x \right )=\frac{2}{32}\left ( 4x^2-x^3/3-12x &plus;32/3 \right );2\leqslant x\leqslant 6}$

The mode is found as

$\frac{\mathrm{d} }{\mathrm{d} x}\rho \left ( x \right )=0\\ k\left ( 8-2x\right )=0\\ {\color{Blue} x_{mode}=4}$

Median is found as

$F\left ( x_{median} \right )=0.5\\ \frac{3}{32}\left ( 4x_{median}^2-x_{median}^3/3-12x_{median}&plus;32/3 \right )=0.5\\ {\color{Blue} x_{median}=4.605}$

We are required to solve only one question and 4 parts. Please post the remaining questions as another post.