Answer to Question #213484 in Statistics and Probability for WWW

Question

An acher shoots an arrow on a target. The distance of the arrow from the center of the target is a random variable x,whose pdf is given by

f(x)={k(3+2x-x²) if x<or equal to 3

{0 if x>3

find the values of the constant k,also find the mean and standard deviation of X

Answer 1

Solution:

We know that $\int_xf(x)=1$

$\Rightarrow \int_0^3k(3+2x-x^2)dx=1 \\ \Rightarrow \int_0^3(3k+2kx-kx^2)dx=1 \\ \Rightarrow [3kx+kx^2-\dfrac{kx^3}{3}]_0^3=1 \\ \Rightarrow [9k+9k-9k]-[0]=1 \\ \Rightarrow 9k=1 \\ \Rightarrow k=\dfrac19$

So, pdf is:

f(x)={$\dfrac19$ (3+2x-x²) if $0\le x\le3$

{0 if x>3

Mean$=E[X]=\int_0^3x.\dfrac19.(3+2x-x^2)dx$

$=\dfrac19\int_0^3(3x+2x^2-x^3)dx \\=\dfrac19[\dfrac{3x^2}{2}+\dfrac{2x^3}{3}-\dfrac{x^4}{4}]_0^3 \\=\dfrac{1}{9}[\dfrac{27}{2}+18-\dfrac{81}{4}-0] \\=\dfrac54$

$E[X^2]=\int_0^3x^2.\dfrac19.(3+2x-x^2)dx \\=\dfrac19\int_0^3(3x^2+2x^3-x^4)dx \\=\dfrac19[x^3+\dfrac{x^4}{2}-\dfrac{x^5}{5}]_0^3 \\=\dfrac{1}{9}[27+\dfrac{81}{2}-\dfrac{243}{5}-0] \\=\dfrac{21}{10}$

Variance$=Var[X]=E[X^2]-(E[X])^2$

$=\dfrac{21}{10}-(\dfrac54)^2=\dfrac{43}{80}$

Standard deviation$=SD[X]=\sqrt{\dfrac{43}{80}}$