Question 16
Bottles of soft drink should contain a volume with a mean of 591 mL, but some variation is expected.
Any bottle at or below the 20th percentile of the volume distribution is rejected. A percentile is a measure in statistics that shows the values below which a given percentage of observations occur.
Thirty-five per cent of the bottles contain 593 mL or more of soft drink.
Assuming the volumes are normally distributed, determine the smallest volume (in mL) that will be accepted.
[4 marks]The 35% fixes the spread, then the 20% sets the cut-off
You know the mean but not the standard deviation. Watch one fact pin down the spread before the rejection line can be drawn.
The 35% above 593 mL is only true for one standard deviation. Find that first, then the 20th percentile has a definite value.
Find the spread that puts 35% above 593 mL
The mean stays at 591 mL. First decide which area goes with 593 mL. Then change the standard deviation and watch both shaded regions move. The rejection line only settles once the blue area is exactly 35%.
Choose the area first. Your calculator’s inverse normal needs an area and which side of the value it sits on.
That gives $z\approx-0.3853$, and then $\sigma=\frac{2}{-0.3853}$ is negative. A standard deviation can’t be negative. 593 mL is above the mean, so the 35% is the area to its right.
Yes. 35% at or above 593 mL leaves 65% below it, so 593 mL is the 65th percentile and $z\approx0.3853$. This choice also sets the slider to the standard deviation that matches.
Now exactly 35% of bottles hold 593 mL or more, so $\sigma\approx5.1905$. The orange cut-off is the 20th percentile, about 586.63 mL. Bottles at or below it are rejected, so the smallest accepted volume is 587 mL to the nearest millilitre.
Less than 35% of bottles reach 593 mL. The curve is too narrow, so increase $\sigma$. Notice the orange cut-off moves too, so you can’t find it until $\sigma$ is right.
More than 35% of bottles reach 593 mL. The curve is too wide, so decrease $\sigma$. The wider the curve, the further the orange cut-off sits below 591 mL.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Your calculator gives both $z$-scores. The work is in knowing which areas to give it, and in what order.
The mean is $\mu=591$, but $\sigma$ is missing. The only other fact about the distribution is that 35% of bottles hold 593 mL or more. That leaves 65% below 593 mL, so 593 mL is the 65th percentile. Use your calculator’s inverse normal function with an area of 0.65 to the left, a mean of 0 and a standard deviation of 1.
$$z\approx0.38532$$
Equivalent decimal values are accepted. Entering a right-tail area of 0.35 gives the same $z$-score. A quick sketch with 591, 593 and both shaded areas helps you check the signs.
Marker: correctly determines the z-score associated with the 65th percentile
The $z$-score says 593 mL is 0.38532 standard deviations above the mean. Put it into the $z$-score formula and solve for $\sigma$.
$$\begin{aligned}z&=\frac{x-\mu}{\sigma}\\0.38532&=\frac{593-591}{\sigma}\\\sigma&=\frac{2}{0.38532}\\&\approx5.1905\end{aligned}$$
If your $z$-score had a slip, you can still earn this mark by follow-through.
Marker: determines $\sigma$
Now the rejection rule. At or below the 20th percentile means an area of 0.2 to the left. Use the inverse normal again with an area of 0.2 to the left, a mean of 0 and a standard deviation of 1.
$$z\approx-0.841621$$
It is negative because the cut-off is below the mean. This mark may be implied by your subsequent working, such as going straight to the inverse normal with a mean of 591 and a standard deviation of 5.1905.
Marker: correctly determines the z-score associated with the 20% rejection region
Let $x$ be the smallest volume that will be accepted. It sits at the boundary of the bottom 20%, so use the $z$-score formula again with the $\sigma$ from step 2.
$$-0.841621=\frac{x-591}{5.1905}$$
$$\begin{aligned}x&=591-0.841621\times5.1905\\&\approx586.63\end{aligned}$$
The 20th percentile is about 586.63 mL, and the sample response rounds it to 587 mL. Label your answer as the smallest volume. Follow-through applies if an earlier step had a slip.
Marker: determines the smallest volume
Each phrase of the question gave you something
There are two normal distribution procedures chained together. You have to notice that the standard deviation is missing, use the 35% fact to find it, and only then find the 20th percentile. The 35% is also a right tail, so you have to convert it to the 65th percentile before you can use it. It is still familiar, because each procedure is one you have practised and the question names the percentile and the mean for you.
If I had to guess where you might lose marks, it would be the first $z$-score. Using 0.35 as a left-tail area gives a negative $z$-score and a negative standard deviation, which should stop you straight away. The effective response shown in the 2020 subject report uses a diagram, states the tail area, mean and standard deviation behind each $z$-score, and labels the answer as the smallest volume. Write those values instead of calculator key presses.
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 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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