2021 Paper 2, Q19
A random variable $X$, defined over the interval $[a,b]$, is uniformly distributed if its probability density function is defined by
$$f(x)=\begin{cases}\dfrac{1}{b-a}, & a\le x\le b\\[4pt] 0, & \text{otherwise}\end{cases}$$
The expected value and variance of a uniform random variable $X$ are
$$E(X)=\frac{a+b}{2},\qquad \text{Var}(X)=\frac{(b-a)^2}{12}$$
A manufacturer has observed that the time that elapses between placing an order with a supplier and the delivery of the order is uniformly distributed between 100 and 180 minutes.
Determine the probability that the time between placing an order and delivery of the order will be within one standard deviation of the expected time.
A uniform distribution, and a band in the middle
Every delivery time between 100 and 180 minutes is equally likely, so there is no peak in the middle. One standard deviation either side of 140 covers 46 of the 80 minutes, so about 58% of orders land inside.
Three formulas are given, and one sentence needs decoding
$a=100$, $b=180$, so $f(x)=\frac{1}{80}$, $E(X)=140$ and $\text{Var}(X)=\frac{80^2}{12}=\frac{1600}{3}$.
The formula gives the variance. You need $\sigma=\sqrt{\frac{1600}{3}}=\frac{40}{\sqrt3}\approx 23.094$ minutes. Forgetting the square root is the most common single error here.
$P\left(140-\frac{40}{\sqrt3}<X<140+\frac{40}{\sqrt3}\right)$. You can find this as an integral of $\frac{1}{80}$, or just as the area of a rectangle.
Change the interval and the answer will not move
Slide the upper end. The mean, the standard deviation and the band all change. But the band always covers $\frac{1}{\sqrt3}$ of the interval, so the probability is stuck at 0.577.
the height is 1 ÷ (b − a), so the area stays 1 as the interval widens
Why is the answer not about 68%?
QCAA marking guide · 4 marks
The width is 80, so the height must be $\frac{1}{80}$ for the total area to equal 1:
$$f(x)=\frac{1}{80},\quad 100\le x\le 180$$
Marker: correctly determines the probability density function.
$$E(X)=\frac{180+100}{2}=140,\qquad \text{Var}(X)=\frac{80^2}{12}=\frac{1600}{3}$$
$$\therefore\ \sigma=\sqrt{\frac{1600}{3}}=\frac{40}{\sqrt3}\approx 23.094$$
Marker: correctly determines the mean and the standard deviation. Exact forms such as $\sqrt{\frac{1600}{3}}$ and decimals such as 23.094 are both accepted.
$$P\left(140-\tfrac{40}{\sqrt3}<X<140+\tfrac{40}{\sqrt3}\right)=\int_{140-40/\sqrt3}^{140+40/\sqrt3}\frac{1}{80}\,dx$$
Marker: establishes a definite integral to represent the probability. QCAA’s Method 2 accepts the graphical version instead. Identify the probability as the area of a rectangle, $\frac{1}{80}\times\frac{80}{\sqrt3}$.
$$=\frac{1}{\sqrt3}=0.57735$$
About a 57.7% chance
Marker: determines the probability. Equivalent decimals accepted, and interim values to at least three decimal places, so 0.576 scores if you carried 23.094.
Every uniform distribution gives 0.577
The band is $2\sigma=\frac{2(b-a)}{2\sqrt3}$ wide out of a total width of $b-a$. The interval cancels, leaving $\frac{1}{\sqrt3}$ whatever $a$ and $b$ are.
The band runs from 116.9 to 163.1 minutes. That is 46.19 of the 80 possible minutes. Deliveries outside it are the fastest 17 minutes and the slowest 17.
0.577 is less than the 0.683 you would get for a normal distribution, and it should be. A flat distribution puts more of its weight out at the edges than a bell curve does.
Everything you need is printed on the page. The density, the mean and the variance are all given, and that is exactly what makes it unfamiliar. There is nothing to look up and nothing to remember, so the question is about what you do with the three formulas. There are two traps. The formula gives you the variance, but the question asks about a standard deviation, so you have to take a square root that nothing tells you to take. And “within one standard deviation of the expected time” is a phrase you have probably only seen with the 68% rule for normal distributions. Here it means a symmetric interval, and you have to work out its probability from scratch. I would expect a lot of students to write 0.68 on autopilot, so be careful!
Question wording and marking-guide steps are from the 2021 QCAA Mathematical Methods 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.