Question 12
The life span of batteries manufactured by a company is assumed to be normally distributed with an unknown mean and standard deviation.
A supervisor at the company randomly selects $n$ batteries and uses the life spans from this sample to calculate an approximate 95% confidence interval for the population mean of $(2321.4,\ 2423.6)$ hours.
Determine the mean life span for this sample of batteries.
[1 mark]The standard deviation of the life spans of batteries in this sample is 125 hours.
Determine $n$.
[3 marks] c)Use the result from Question 12b) to explain whether the assumption that the life span of batteries is normally distributed is required to support the supervisor’s calculations.
[2 marks]Reading an interval backwards
The interval is centred on the sample mean, and half its width is the margin of error. Rearranging the formula for the margin of error then gives the sample size.
The orange interval is the supervisor’s. The blue interval uses the same sample mean and $s=125$, with other sample sizes.
Find the sample size, then test the shape
For part b), move the sample size until your interval matches the supervisor’s. For part c), keep the sample size and change the shape of the life spans. The black curve shows how the mean of $n$ batteries is distributed.
This interval is wider than the supervisor’s, so this sample is too small. Fewer batteries means a bigger margin of error. Try a larger $n$.
This interval is narrower than the supervisor’s, so this sample is too big. More batteries means a smaller margin of error. Try a smaller $n$.
This matches the supervisor’s interval to one decimal place, so the supervisor used $n=23$ batteries. Its neighbours miss: $n=22$ gives a margin of error of about $52.2$ and $n=24$ about $50.0$.
When the life spans are normal, the mean of any number of them is exactly normal, so the two curves sit on top of each other for every sample size. This is what the supervisor’s calculation relies on.
With skewed life spans and fewer than 30 batteries, the mean keeps some of that skew, and the smaller the sample, the more skew it keeps. So the mean cannot be assumed to be normal, and the interval would not be justified. The question’s sample of $n=23$ is in this range, so the assumption is required.
With $n\ge30$, the central limit theorem says the mean is approximately normal whatever the shape of the life spans. In this course that counts as large enough, so the assumption would not be needed.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide.
A confidence interval for the mean is built as $\bar{x}\pm E$, so the sample mean is the midpoint of the interval.
$$\bar{x}=\frac{2321.4+2423.6}{2}$$
Marker: correctly determines the mean lifetime of the sample
The formula sheet gives the approximate confidence interval for $\mu$. The question gives the sample standard deviation, $s=125$, and for 95% your GDC gives $z\approx1.96$.
$$\left(\bar{x}-z\frac{s}{\sqrt{n}},\ \bar{x}+z\frac{s}{\sqrt{n}}\right)$$
The upper end of the interval gives an equation with $n$ as the only unknown.
$$2372.5+1.96\times\frac{125}{\sqrt{n}}=2423.6$$
The QCAA accept equivalent equations, such as $1.96\times\dfrac{125}{\sqrt{n}}=51.1$ or $2372.5-2\left(\dfrac{125}{\sqrt{n}}\right)=2321.4$, and follow-through applies from part a). Be aware that the second one uses $z=2$ and leads to $n\approx23.94$, so $n=24$. On Paper 2, use $1.96$ from your GDC, as the sample response does.
Marker: determines an equation in terms of $n$
Solve it. You can use your GDC’s solver, or rearrange by hand. Remember to square at the end.
$$\begin{aligned}1.96\times\frac{125}{\sqrt{n}}&=51.1\\\sqrt{n}&=\frac{245}{51.1}\approx4.795\\n&\approx22.99\end{aligned}$$
The QCAA also accept $\sqrt{n}\approx4.79$, and this mark can be implied by your next line. The marking guide prints $n\approx22.98$; the exact value is $22.987\ldots$, which rounds to $22.99$. Either way, the next step is the same.
Marker: solves equation for $n$
$n$ counts batteries, so it must be a whole number. Here you are recovering the sample size the supervisor actually used, so round to the nearest whole number. Game plan 9.5 says to round up, but that is for questions asking for the smallest sample that achieves a given precision. Here both give the same answer.
Check it: $1.96\times\dfrac{125}{\sqrt{23}}\approx51.09$, and $2372.5\pm51.09$ gives $(2321.4,\ 2423.6)$ to one decimal place.
Marker: expresses sample size rounded to a whole number
The interval treats the sample mean as normally distributed. Start with a clear decision.
Yes, the assumption is required.
If your part b) gave $n\ge30$, the QCAA allow a follow-through mark for answering no.
Marker: makes a suitable statement
Now give the reason, using your $n$ from part b). The central limit theorem only makes the sample mean approximately normal for large samples, which in this course means $n\ge30$.
The QCAA accept equivalent explanations, such as “the sample size is small, so the assumption of normality is required”. If part b) gave $n\ge30$, a follow-through mark is allowed for explaining that the sample is large enough.
Marker: discusses the sample size and its relation to a normal distribution
Each phrase of the question gave you something
Working backwards from an interval is a routine the QCAA use often. The midpoint and the half-width come straight off the interval, and the margin of error formula then gives $n$. Part c) then checks the $n\ge30$ condition on the $n$ you have just found.
If I had to guess where you might lose marks, it would be part c). “Yes” on its own earns only one mark. The second mark needs the reason: $n=23$ is under 30, so the sample mean is only normal if the life spans are. In part b), use half the width, $51.1$, and remember to square $\sqrt{n}$. Using the full width of $102.2$ gives $n\approx6$.
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.
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