Question 19
Consider the following information.
| Continuous random variable $X$ | mean | $E(X)=\mu=\int_{-\infty}^{\infty}x\,p(x)\,dx$ |
| variance | $\text{Var}(X)=\int_{-\infty}^{\infty}(x-\mu)^{2}p(x)\,dx$ |
The waiting time (minutes) until workers at a certain call centre receive their $n$th phone call, where $n\in\mathbb{Z}^{+}$, is a random variable $T$ with probability density function
$$f(t)=\begin{cases}\dfrac{k^{n}t^{n-1}}{(n-1)!}e^{-\frac{t}{3}}, & t\ge0\\[4pt]0, & \text{otherwise}\end{cases}$$
where $k$ is a positive constant.
The waiting time until workers receive their 5th call is collected from a random sample of 80 workers.
Determine the probability that the mean waiting time from this sample is more than 16 minutes.
From one skewed wait to a narrow normal curve
Watch the population curve for one worker, then the curve for the mean of 80 workers. The question is about the second one.
One worker waiting more than 16 minutes is common. The average of 80 workers being over 16 minutes is rare.
See how the sample size changes the probability
$\bar{T}$ has mean 15 and standard deviation $\frac{\sqrt{45}}{\sqrt{n}}$. Change the sample size and the cut-off to see how the shaded probability changes.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The first four marks are about the population, and the last three are about the sample mean.
The area under a probability density function is 1. Put $n=5$ into the formula, so $(n-1)!=4!$.
$$\int_{0}^{\infty}\frac{k^{5}t^{4}}{4!}e^{-\frac{t}{3}}\,dt=1$$
Any value of $n$ works for finding $k$, such as $n=1$, but you need $n=5$ later anyway.
Marker: correctly determines an equation in terms of $k$
Solve it on your calculator.
$$k=\frac{1}{3}$$
You might see it as $243k^{5}=1$ first, which is also accepted.
Marker: solves the equation to determine $k$
Use the mean formula from the table, with $k=\frac{1}{3}$.
$$\mu=\int_{0}^{\infty}t\cdot\frac{\left(\frac{1}{3}\right)^{5}t^{4}}{4!}e^{-\frac{t}{3}}\,dt=15\text{ minutes}$$
You keep follow-through marks from here if you slipped earlier.
Marker: determines the population mean
Use the variance formula from the table.
$$\sigma^{2}=\int_{0}^{\infty}(t-15)^{2}\frac{\left(\frac{1}{3}\right)^{5}t^{4}}{4!}e^{-\frac{t}{3}}\,dt=45\qquad\sigma=\sqrt{45}\approx6.71$$
Marker: determines the population variance
The question is about the mean of the sample, not one worker. Let $\bar{T}$ be the sample mean. You need to say why it is normal, because the population curve is not.
The sample size of 80 is large ($n\ge30$), so by the central limit theorem $\bar{T}$ is approximately normally distributed.
Marker: justifies that the distribution of $\bar{T}$ can be considered normal
The sample mean has the same mean as the population, and its standard deviation is divided by $\sqrt{n}$.
$$\mu_{\bar{T}}=15\qquad\sigma_{\bar{T}}=\frac{\sqrt{45}}{\sqrt{80}}=0.75$$
Marker: determines the mean and standard deviation of the sample mean
Use normal cdf on your calculator with a lower bound of 16, an upper bound of $\infty$, $\mu=15$ and $\sigma=0.75$.
$P\left(\bar{T}>16\right)\approx0.09$
You can also give it as about 9% or 0.1. Using $\sigma=\sqrt{45}$ instead of $0.75$ here is the most common way to lose this mark.
Marker: determines the required probability
Each phrase of the question gave you something
The density function is not one you have met before, and it has an unknown constant and a general $n$ in it. You have to find $k$, the mean and the variance yourself before you can use any of the sample mean ideas.
The line I would check first is the standard deviation you use at the end. The QCAA often give you a population and then ask about a sample mean, so you need to divide by $\sqrt{n}$ before you find the probability.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.