STEM Problem Detail

Problem #90

Information and Communications Technology (ICT) Hard

Problem Statement

Prompt

Give the solution of Example 6

Solution Rendering...

\[
\int_0^1 f(x)\,dx = 1
\]

\[
k\int_0^1 (x-x^2)\,dx = 1
\]

\[
k\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = 1
\]

\[
k\left(\frac{1}{2}-\frac{1}{3}\right)=1
\;\Rightarrow\;
\frac{k}{6}=1
\;\Rightarrow\;
k=6
\]

\[
\text{A.M.}=E(X)=\int_0^1 x\cdot 6x(1-x)\,dx
=6\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1
=\frac{1}{2}
\]

\[
\text{H.M.}=\left[\int_0^1 \frac{1}{x}\,6x(1-x)\,dx\right]^{-1}
=\left[\int_0^1 6(1-x)\,dx\right]^{-1}
=\frac{1}{3}
\]

\[
\int_0^M 6x(1-x)\,dx=\frac{1}{2}
\]

\[
(3x^2-2x^3)_0^M=\frac{1}{2}
\]

\[
2M^3-3M^2+\frac{1}{2}=0
\]

\[
M=\frac{1}{2}
\]

\[
f(x)=6x(1-x)
\]

\[
f'(x)=6-12x,\qquad f''(x)=-12
\]

\[
f'(x)=0 \Rightarrow x=\frac{1}{2},\quad f''(x)<0
\]

\[
\text{Mode}=\frac{1}{2}
\]

\[
E(X^2)=\int_0^1 6x^2(1-x)\,dx
=6\left[\frac{x^4}{4}-\frac{x^5}{5}\right]_0^1
=\frac{3}{10}
\]

\[
\operatorname{Var}(X)=E(X^2)-[E(X)]^2
=\frac{3}{10}-\frac{1}{4}
=\frac{1}{20}
\]

\[
\text{S.D.}=\sqrt{\frac{1}{20}}=\frac{1}{2\sqrt{5}}
\]

\[
\text{M.D. about mean}
=E\!\left|X-\frac{1}{2}\right|
=\int_0^1 \left|x-\frac{1}{2}\right|6x(1-x)\,dx
\]

\[
=\int_0^{1/2}\!\!\left(\frac{1}{2}-x\right)6x(1-x)\,dx
+\int_{1/2}^{1}\!\!\left(x-\frac{1}{2}\right)6x(1-x)\,dx
=\frac{3}{16}
\]

\[
\mu_3=E\!\left(X-\frac{1}{2}\right)^3=0
\]

\[
\mu_4=E\!\left(X-\frac{1}{2}\right)^4
=6\int_0^1 \left(x-\frac{1}{2}\right)^4 x(1-x)\,dx
=\frac{3}{560}
\]

\[
\beta_1=\frac{\mu_3^2}{\mu_2^3}=0
\]

\[
\beta_2=\frac{\mu_4}{\mu_2^2}
=\frac{\tfrac{3}{560}}{(\tfrac{1}{20})^2}
=\frac{15}{7}
\]

\[
\therefore\ \text{the distribution is symmetric and platykurtic.}
\]

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