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Calculus, SPEC1 2025 VCAA 4

The waiting time, \(T\) hours, to see a particular doctor at a clinic has a distribution with a probability density function \(f\) defined by

\begin{align*}
f(t)=\left\{\begin{array}{cl}
\dfrac{3}{2 \log _e(2)}\left(\dfrac{1}{(t+1)(2-t)}\right), & 0<t \leq 1 \\
0, & \text{elsewhere}
\end{array}\right.
\end{align*}

  1. Use integration to show that  \(E (T)=\dfrac{1}{2}\).   (3 marks)

  2. For random samples of 25 waiting times, it may be assumed that the sample means are approximately normally distributed.
  3. Find the probability that the average waiting time for a random sample of 25 patients is between 0.44 hours and 0.5 hours.
  4. Use  \(\sigma=0.3\)  and  \(\operatorname{Pr}(Z<1)=0.84\)   (2 marks)

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a.    \(\text{See Worked Solutions}\)

b.    \(\operatorname{Pr}(0.44<\overline{T}<0.5)=0.34\)

Show Worked Solution

a.    \(E(T)=\dfrac{3}{2 \ln 2} \displaystyle \int_0^1 \dfrac{t}{(t+1)(2-t)}\, dt\)
 

\(\text {Using partial fractions:}\)

\(\dfrac{t}{(t+1)(2-t)}=\dfrac{A}{(t+1)}+\dfrac{B}{(2-t)}\)

\(A(2-t)+B(t+1)=t\)

\(\text{If}\ \ t=2:\)

\(3 B=2 \ \Rightarrow \ B=\dfrac{2}{3}\)

\(\text{If}\ \ t=-1:\)

\(3 A=-1 \ \Rightarrow \ A=-\dfrac{1}{3}\)

Mean mark (a) 51%.
\(E(T)\) \(=\displaystyle \frac{3}{2 \ln 2} \int_0^1-\frac{1}{3(t+1)}+\frac{2}{3(2-t)} d t\)
  \(=\displaystyle\frac{1}{2 \ln 2} \int_0^1-\frac{1}{t+1}+\frac{2}{2-t} d t\)
  \(=\displaystyle\frac{1}{2 \ln 2}\Big[-\ln \abs{t+1}+2 \ln \abs{2-t}\Big]_0^1\)
  \(=\displaystyle\frac{1}{2 \ln 2}[(-\ln 2+2 \ln 1)-(2 \ln 1-2 \ln 2)]\)
  \(=\displaystyle\frac{1}{2 \ln 2}(\ln 2)\)
  \(=\displaystyle\frac{1}{2}\)

 

b.    \(E(\overline{T})=\dfrac{1}{2}, \ \operatorname{sd}(\overline{T})=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.3}{\sqrt{25}}=0.06\)

\(\operatorname{Pr}(0.44<\overline{T}<0.5)\) \(=\operatorname{Pr}\left(\dfrac{0.44-0.5}{0.06}<Z<0\right)\)
  \(=\operatorname{Pr}(-1<Z<0)\)
  \(=\operatorname{Pr}(0<Z<1)\)
  \(=0.84-0.50\)
  \(=0.34\)