Answer to Question #104663 in Statistics and Probability for Amra Musharraf

Question

Calculate the standard deviation and mean deviation from mean if the frequency function f(x) has the form: f(x)={3+2x/18,for2≤x≤4; 0 otherwise

Answer 1

The frequency function f(x) has the form:

$f(x) = \begin{cases} {3+2x \over 18}, &\text{for } 2\leq x\leq4 \\ 0, &\text{otherwise } \end{cases}$

$\displaystyle\int_{-\infin}^\infin f(x)dx=\displaystyle\int_{2}^4 {3+2x \over 18}dx=\bigg[{3x+x^2 \over 18}\bigg]\begin{matrix} 4 \\ 2 \end{matrix}=$ $={3(4)+(4)^2 \over 18}-{3(2)+(2)^2 \over 18}=1$

$\mu=E(X)=\displaystyle\int_{-\infin}^\infin xf(x)dx=\displaystyle\int_{2}^4 x{3+2x \over 18}dx=$

$=\bigg[{{3x^2 \over 2}+{2x^3 \over 3} \over 18}\bigg]\begin{matrix} 4 \\ 2 \end{matrix}={9(4)^2+4(4)^3 \over 108}-{9(2)^2+4(2)^3 \over 108}={83 \over 27}$

$E(X^2)=\displaystyle\int_{-\infin}^\infin x^2f(x)dx=\displaystyle\int_{2}^4 x^2{3+2x \over 18}dx=$

$=\bigg[{{3x^3 \over 3}+{2x^4 \over 4} \over 18}\bigg]\begin{matrix} 4 \\ 2 \end{matrix}={2(4)^3+(4)^4 \over 36}-{2(2)^3+(2)^4 \over 36}={88 \over 9}$

$Var(X)=\sigma^2=E(X^2)-(E(X))^2=$

$={88 \over 9}-\big({83 \over 27}\big)^2={239 \over 729}$

$\mu=E(X)={83 \over 27}$

$Var(X)=\sigma^2={239 \over 729}$

$\sigma_X=\sqrt{Var(X)}=\sqrt{{239 \over 729}}\approx0.5726$

$MD=\displaystyle\int_{-\infin}^\infin (x-\mu)f(x)dx=$

$=\displaystyle\int_{2}^4 (x-{83 \over 27}){3+2x \over 18}dx=$

$=\bigg[{36x^3-85x^2-498x\over 972}\bigg]\begin{matrix} 4 \\ 2 \end{matrix}=0$