Question 13
Data records show that the speeds of cars at a particular location on a highway are normally distributed with a mean of $98.7\text{ km h}^{-1}$ and a standard deviation of $4.1\text{ km h}^{-1}$. The speed limit at this location is $100\text{ km h}^{-1}$.
A police officer plans to record the speeds of 20 randomly selected cars at this location.
Determine the expected number of cars in the sample that will be travelling within $\pm\,1\text{ km h}^{-1}$ of the population mean.
[2 marks] b)Determine the probability that the mean speed of this sample will exceed the speed limit.
[2 marks]There is a 5% probability that the mean speed of this sample will exceed $k$.
Determine the value of $k$.
[2 marks]One car’s speed varies far more than the mean of 20 cars
Part a) is about single cars, so it uses 4.1. Parts b) and c) are about the mean speed of the officer’s 20 cars, so they use 4.1 ÷ √20.
About 38% of single cars go faster than 100, but the mean of 20 cars does so only about 8% of the time.
Change the band, the sample size and the tail
Pick a part, then move its slider. Part a) widens or narrows the band around the mean. Part b) changes the number of cars in the sample. Part c) changes the size of the top tail.
This is the question’s band, within $1\text{ km h}^{-1}$ of the mean. The probability is for one car, so the standard deviation is 4.1. Multiply by 20 to get the part a) answer.
This band is narrower than the question’s, so it holds less of the curve and fewer of the 20 cars are expected in it.
This band is wider than the question’s, so it holds more of the curve and more of the 20 cars are expected in it.
This is the officer’s sample of 20 cars, so this is the part b) answer. The speeds are normal, so the sample mean is normal too, even though 20 is less than 30.
This is one car on its own, so the curve is the population curve. Using 4.1 as the standard deviation in part b) gives this answer, which is the chance for one car.
A smaller sample gives a wider curve, so more of it lies above the speed limit.
A bigger sample gives a narrower curve, so less of it lies above the speed limit.
This is the question’s 5%, so this is the part c) answer. The tail is smaller than the 7.8% above 100 from part b), so $k$ sits just above the speed limit.
This tail is smaller than the 7.8% above 100 from part b), so $k$ sits above the speed limit.
This tail is bigger than the 7.8% above 100 from part b), so $k$ sits below the speed limit.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide. This is a Technology-active question, so the probabilities and the inverse normal come from your GDC.
Part a) is about the speeds of individual cars, so use the population distribution. Let $X$ be the speed of one car.
$$X\sim N(98.7,\ 4.1^2)$$
Within $\pm\,1$ of the mean is from 97.7 to 99.7. Use the normal cumulative distribution on your GDC with those bounds, a mean of 98.7 and a standard deviation of 4.1.
$$P(97.7\le X\le99.7)\approx0.193$$
This mark can be implied by your next line.
Marker: correctly calculates required probability
Each of the 20 cars has the same chance of being in the band, so the expected number is 20 times that chance.
$$\begin{aligned}&\text{Expected number}\\&\quad=20\times0.1927\\&\quad\approx3.85\end{aligned}$$
Multiply by the unrounded probability from your GDC. The marking guide prints $20\times0.193$ beside 3.85, but 3.85 comes from the unrounded value, because $20\times0.193=3.86$. The guide then rounds to about 4. An expected number is an average over many samples, so it doesn’t have to be a whole number. If your probability was wrong, you can still earn this mark for multiplying it by 20.
Marker: calculates the expected number
Part b) is about the mean speed of the sample, so you need the distribution of $\bar{X}$. The speeds are normally distributed, so $\bar{X}$ is normal for any sample size, even though $n=20$ is less than 30. Its mean is the population mean, and its standard deviation is $\sigma$ divided by $\sqrt{n}$.
$$\mu_{\bar{X}}=98.7$$
$$\begin{aligned}\sigma_{\bar{X}}&=\frac{\sigma}{\sqrt{n}}=\frac{4.1}{\sqrt{20}}\\&\approx0.917\end{aligned}$$
This mark can be implied by your next line.
Marker: correctly calculates $\sigma_{\bar{X}}$
Use the normal cumulative distribution with a lower bound of 100, a very large upper bound such as $10^{99}$, a mean of 98.7 and a standard deviation of $\frac{4.1}{\sqrt{20}}$. Enter the standard deviation as that exact expression so nothing is lost to rounding.
The QCAA also accept more decimal places, such as $0.078$. If your $\sigma_{\bar{X}}$ was wrong, you can still earn this mark for using it correctly.
Marker: calculates required probability
Part c) is about the same sample mean, so keep $\sigma_{\bar{X}}=\frac{4.1}{\sqrt{20}}$. A 5% chance of exceeding $k$ is a right tail of area 0.05. Show it with a sketch or with a probability statement.
$$P(\bar{X}\gt k)=0.05$$
This is the same as
$$P(\bar{X}\le k)=0.95$$
Marker: correctly uses appropriate mathematical representation
Use the inverse normal on your GDC with an area of 0.95 to the left, a mean of 98.7 and a standard deviation of $\frac{4.1}{\sqrt{20}}$. If your GDC asks for a tail, you can choose the right tail and enter an area of 0.05 instead.
The QCAA also accept $100.21$, and whole-number answers of 100 or 101. Check that $k$ is above the mean. A top 5% has to be.
Marker: correctly determines the value of $k$
Each phrase of the question gave you something
Every part is a GDC calculation you have done many times: a normal probability, an expected number, a probability for a sample mean and an inverse normal. The only decision is which standard deviation to use. Part a) is about single cars and uses 4.1. Parts b) and c) are about the mean of 20 cars and use $\frac{4.1}{\sqrt{20}}$.
If I had to guess where you might lose marks, it would be that choice of standard deviation. Using $\frac{4.1}{\sqrt{20}}$ in part a) gives a probability of about 0.725 and about 14.5 cars, which is far too many for such a narrow band of speeds. Using 4.1 in part b) gives about 0.376, the chance for a single car. In part c), entering 0.05 as the area to the left gives $k\approx97.2$, which is below the mean and so can’t be the start of the top 5%.
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 Specialist Mathematics 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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