smc-7137-45-\(E(X)/\text{Var}(X)\)

A continuous random variable \(X\) has probability density function \(f(x)\) given by

\begin{align*}
f(x)=\left\{\begin{array}{cl}
k x(1-x)^5, & \text { for } 0 \leq x \leq 1 \\
0, & \text { for all other values of } x
\end{array}\ \ \ , \text { where } k\right. \text { is a constant. }
\end{align*}

It is given that

\(\displaystyle \int_0^a x(1-x)^5\, d x=\frac{1}{42}+\frac{(1-a)^7}{7}-\frac{(1-a)^6}{6}\)

and \(\displaystyle\int_0^1 x^m(1-x)^5\, d x=\dfrac{120}{(m+1)(m+2)(m+3)(m+4)(m+5)(m+6)}\)

where \(a>0\) and \(m>0\).

Show that \(k=42\). (1 mark)

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Show that \(E (X)=0.25\). (2 marks)

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Show that the median of \(X\) is less than the expected value of \(X\). (3 marks)

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a. \(k \displaystyle \int_0^1 x(1-x)^5 d x=1\)

\(k\left[\dfrac{1}{42}+\dfrac{(1-1)^7}{7}-\dfrac{(1-1)^6}{6}\right]=1\)

\(\dfrac{k}{42}\)	\(=1\)
\(k\)	\(=42\)

b.	\(E (X)\)	\(=\displaystyle \int_0^1 x \times f(x)\, d x\)
		\(=\displaystyle \int_0^1 42 x^2(1-x)^5\, d x\)
		\(=42 \times \dfrac{120}{3 \times 4 \times 5 \times 6 \times 7 \times 8}\)
		\(=0.25\)

c. \(\text{Let}\ m =\text{ median}\)

\(P(X\leqslant m) = 0.5\ \ \Rightarrow\ \ \displaystyle \int_0^m 42 x(1-x)^5\, d x=0.5 \)

\(E(X)=0.25\)

\(\text{Calculate }\ P(X\leqslant 0.25):\)

\(\displaystyle \int_0^{0.25} 42 x(1-x)^5\, d x\)	\(=42\left[\frac{1}{42}+\dfrac{(1-0.25)^7}{7}-\dfrac{(1-0.25)^6}{6}\right]\)
	\(=0.555 \ldots\ \text{(3 dp)}\)

\(\therefore \displaystyle \int_0^m 42 x(1-x)^5 d x=0.5 \ \ \text{requires the median to be less than 0.25.}\)

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