Question 18
A random variable $X$ is normally distributed, with a known mean $\mu$ and standard deviation $\sigma$.
In figure 1, the shaded region between 4 and $\mu$ represents 30% of the distribution of $X$.
Consider the distribution of $\bar{X}$ based on repeated random sampling of $X$ using a certain sample size.
In figure 2, the shaded region between $\mu$ and 6 represents 30% of the distribution of $\bar{X}$.
Given $P\left(4\le\bar{X}\le6\right)\approx0.77$, determine $P(4\le X\le6)$.
Two curves share the same centre but not the same spread
Watch how the gap from 4 to $\mu$ is measured on each curve. You will see that comparing the two gaps is what gives you the sample size.
The drawing uses the values of $\mu$ and $\sigma$ that fit both figures, so you can see the real shapes. You never need to find them yourself.
Find the sample size that makes both figures true
You are never told $\mu$ or $\sigma$, but the two figures pin them down. Here they are fixed at those values, so you can change $n$ and see which sample size gives you both 30% and 77%.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This follows the marking guide's z-score method, and you can round the z-values a little differently without losing marks.
Figure 1 tells you 30% of $X$ lies between 4 and $\mu$. Half of any normal distribution is below $\mu$, so 20% of $X$ lies below 4. You use inverse normal on the standard normal to turn that into a z-score.
$$P(Z<z)=0.2\quad\Rightarrow\quad z\approx-0.8416$$
$$\frac{4-\mu}{\sigma}=-0.8416\quad\ldots(1)$$
You can also write this as $\mu=4+0.84\sigma$. This mark can be implied by your later working.
Marker: correctly determines an equation from Figure 1 in terms of $\mu$ and $\sigma$
Figure 2 is about $\bar{X}$, so you need its distribution. Call the sample size $n$. You know the sample mean has the same mean as $X$, but its standard deviation is $\frac{\sigma}{\sqrt{n}}$.
$$\bar{X}\sim N\left(\mu,\left(\frac{\sigma}{\sqrt{n}}\right)^{2}\right)$$
30% of $\bar{X}$ lies between $\mu$ and 6, so 80% lies below 6, and inverse normal gives you $z\approx0.8416$.
$$\frac{6-\mu}{\sigma/\sqrt{n}}=0.8416\quad\ldots(2)$$
Marker: correctly determines an equation from Figure 2 in terms of $\mu$, $\sigma$ and $n$
Now you use the probability you were given. The 0.77 covers both sides of $\mu$, so you take off the 30% from figure 2 to get the part between 4 and $\mu$.
$$P\left(4\le\bar{X}\le\mu\right)\approx0.77-0.3=0.47$$
That leaves $0.5-0.47=0.03$ of $\bar{X}$ below 4, and inverse normal gives you $z\approx-1.8808$.
$$\frac{\mu-4}{\sigma/\sqrt{n}}=1.8808\quad\ldots(3)$$
Marker: correctly determines an equation using $P\left(4\le\bar{X}\le6\right)\approx0.77$ in terms of $\mu$, $\sigma$ and $n$
Equations (1) and (3) both describe the same gap, from 4 up to $\mu$. You measure it once in units of $\sigma$ and once in units of $\frac{\sigma}{\sqrt{n}}$. Rearrange both for $\mu-4$ and set them equal.
$$\mu-4=0.8416\sigma=\frac{1.8808\sigma}{\sqrt{n}}$$
The $\sigma$ cancels, so you can solve for $n$ without ever knowing $\mu$ or $\sigma$.
$$\sqrt{n}=\frac{1.8808}{0.8416}\approx2.235\quad\Rightarrow\quad n\approx5$$
You keep follow-through marks here if you slipped earlier, and you can give $\sqrt{n}\approx2.23$.
Marker: determines $n$
You already have the z-score of 4 for $X$ from (1). Now you need the z-score of 6 for $X$. Multiply (2) through by $\frac{1}{\sqrt{n}}$ to change it from $\bar{X}$ units back to $X$ units.
$$\frac{6-\mu}{\sigma}=\frac{0.8416}{\sqrt{n}}\approx\frac{0.8416}{2.235}\approx0.3766$$
So the probability you want sits between two z-scores.
$$P(4\le X\le6)=P(-0.8416\le Z\le0.3766)$$
You can round these to $-0.84$ and $0.38$.
Marker: expresses required probability in terms of $z$-scores
Use normal CDF on your calculator with the standard normal.
$$P(-0.8416\le Z\le0.3766)\approx0.45$$
$P(4\le X\le6)\approx0.45$
Rounding to 0.4 or 0.5 is also accepted. Notice that you never found $\mu$ or $\sigma$. You only needed how far 4 and 6 are from $\mu$, measured in standard deviations.
Marker: determines required probability
Each phrase of the question gave you something
You are not given $\mu$, $\sigma$ or the sample size. You have to work with z-scores from start to finish, and you have to link two different distributions that share the same mean.
The step I think most of you will find hardest is step 4, because you have to see that two equations describe the same gap. The QCAA like to give you a probability that covers both sides of the mean, so you have to split the 0.77 before you can use inverse normal.
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 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.