Question 3
The masses of packages of cheese produced by a company are assumed to be normally distributed with a known mean of $\mu$ grams and a standard deviation of $7.37$ grams.
The packages of cheese are labelled to contain $500$ grams.
Given there is a $25\%$ probability that the mean mass of $20$ randomly selected packages will be less than the labelled amount, the value of $\mu$ is
Not quite. $498.89$ has the sign of $z$ flipped: $500-0.6745\times1.648\approx498.89$. With the mean below $500$, about $75\%$ of sample means would be below the label. For only $25\%$ below, $500$ must sit below the mean, so $\mu>500$.
Not quite. $500.25$ divides by $n$ instead of $\sqrt n$: $\frac{7.37}{20}\approx0.3685$ and $500+0.6745\times0.3685\approx500.25$. The standard deviation of the sample mean is $\frac{\sigma}{\sqrt n}=\frac{7.37}{\sqrt{20}}\approx1.648$.
That’s it. Step through the hints below to see why.
Not quite. $504.98$ uses $\sigma=7.37$ as the spread, which is for one package: $500+0.6745\times7.37\approx504.97$, which is (D) to within rounding. The mean of $20$ packages varies less, with standard deviation $\frac{7.37}{\sqrt{20}}\approx1.648$.
Mass in grams. The shaded area is the probability of being below the labelled $500$ g. Each spread is drawn to the same height, so only the width changes when you switch.
A quarter of the area is below $500$. With the spread of the sample mean, the mean mass is $\mu\approx501.11$ grams. This is option (C).
With the mean below $500$, three quarters of the area is below the label. For only a quarter below, $500$ has to sit below the mean.
More than a quarter of the area is below $500$. Move the mean up.
Less than a quarter of the area is below $500$. Move the mean down.
A quarter of this curve is below $500$, but the spread $7.37$ is for one package, $X$. This is where (D) comes from. The question is about the mean of $20$ packages.
A quarter of this curve is below $500$, but the spread divides $7.37$ by $20$ instead of $\sqrt{20}$. This is where (B) comes from.
The spread $7.37$ is for one package, $X$. The question is about the mean of $20$ packages, which has the narrower spread $\frac{7.37}{\sqrt{20}}$.
Dividing by $20$ instead of $\sqrt{20}$ makes the curve far too narrow. The spread of the sample mean is $\frac{7.37}{\sqrt{20}}$.
The question is about the mean of $20$ packages, so use the distribution of $\bar{X}$. The masses are normal, so $\bar{X}$ is normal too, with standard deviation $\frac{\sigma}{\sqrt n}$.
$$\bar{X}\sim N\left(\mu,\ \left(\frac{7.37}{\sqrt{20}}\right)^2\right)$$
$$\frac{7.37}{\sqrt{20}}\approx1.648$$
Only $25\%$ of sample means are below $500$, so $500$ is below the mean and its $z$-score is negative. Use inverse normal on your GDC with area $0.25$.
$$P\left(\bar{X}<500\right)=0.25$$
$$z\approx-0.6745$$
Set the $z$-score of $500$ equal to this value and solve for $\mu$.
$$\frac{500-\mu}{1.648}=-0.6745$$
$$\mu\approx500+0.6745\times1.648\approx501.11$$
For samples of $20$, the sample mean is normal with standard deviation $\frac{7.37}{\sqrt{20}}\approx1.648$.
$$P\left(\bar{X}<500\right)=0.25$$
$$\frac{500-\mu}{1.648}\approx-0.6745$$
$$\mu\approx500+0.6745\times1.648\approx501.11$$
The value of $\mu$ is about $501.11$ grams.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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