Question 14
The heights of students at School A are normally distributed with a mean of 165 cm and a standard deviation of 15 cm.
Determine the probability that a student chosen at random from School A is shorter than 180 cm.
[1 mark] b)Determine the minimum integer value of the height of a student who is in the top 2% of this distribution.
[3 marks]The heights of students at School B are also normally distributed. A student at School B has the same height as the height determined in Question 14b) but their corresponding $z$-score is 3.
Determine which student’s height ranks higher in terms of percentile for their school.
[3 marks]One bell curve, three questions about it
Part a) asks for an area, part b) runs it backwards from an area to a height, and part c) puts the height on the $z$-score scale that every normal distribution shares.
Both schools are normal, so a percentile depends only on the $z$-score. A bigger $z$-score means a higher percentile.
Find the smallest whole height in the top 2%
Part b) turns on one decision: which whole number counts as the minimum. Slide the height and watch the orange area, the proportion of School A students who are taller. A height is in the top 2% when that area is 0.02 or less.
This is part a). 180 cm is one standard deviation above the mean, and $P(X<180)\approx0.841$. The orange area, about 0.159, is everyone taller, far more than 2%.
Exactly 2% of School A is taller than 195.806 cm, so this is where the top 2% starts. It isn’t a whole number, though. Try the whole numbers either side of it.
Not in the top 2%. About 2.28% of students are taller than 195 cm, which is more than 2%. Rounding the cut-off down puts you just outside.
In the top 2%: only about 1.94% of students are taller. 195 cm misses out, so 196 cm is the minimum integer height. For part c), its $z$-score is $\frac{31}{15}\approx2.067$, less than School B’s 3, so the School B student ranks higher.
In the top 2%, but not the minimum. 196 cm is shorter and still makes it.
About 1.94% of students are shorter than 134 cm. This is the bottom 2%, close to the 134.194 cm you get if you enter 0.02 as the area to the left. The top 2% is at the other end of the curve.
More than 2% of School A is taller than this, so it isn’t in the top 2%. Move the height up.
Heights here are whole numbers, except the cut-off button. Your calculator’s inverse normal function gives the cut-off from an area to the left.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Parts a) and b) are one calculator command each, with a mean of 165 and a standard deviation of 15. Part c) is a comparison you set out in words.
“Shorter than 180 cm” is the area to the left of 180. Use your calculator’s normal cumulative probability function with an upper bound of 180, a lower bound far below the mean (such as 0), a mean of 165 and a standard deviation of 15.
$$\begin{aligned}P(X<180)&=0.84134\ldots\\&\approx0.841\end{aligned}$$
Equivalent decimal values are accepted. A quick sense check: 180 is one standard deviation above the mean, and about 84% of a normal distribution sits below that.
Marker: correctly determines the probability
Sketch it before you calculate. The top 2% is the upper tail, so the height $x$ you want has an area of 0.98 to its left. The sample response draws exactly this: a bell curve with the area left of $x$ shaded.
The guide accepts equivalent statements or representations, for example finding the upper limit of the interval with area 0.98, mean 165 cm and lower limit 0.
Marker: correctly uses an appropriate mathematical representation
Use your calculator’s inverse normal function with an area of 0.98 to the left, a mean of 165 and a standard deviation of 15.
$$x=195.806\ \text{cm}$$
This mark may be implied by subsequent working.
Marker: correctly determines the lowest decimal height
Anyone at 195.806 cm or taller is in the top 2%. The smallest whole number at or above that is 196.
Follow-through applies if your decimal height had a slip. Here 195.806 rounds to 196 either way, but for a minimum always round up. A check of our own: about 2.28% of students are taller than 195 cm, so 195 is not in the top 2%.
Marker: determines lowest whole height
You know the School B student’s $z$-score. To compare, find the School A student’s $z$-score for 196 cm.
$$\begin{aligned}z_{\text{School A}}&=\frac{196-165}{15}\\&=\frac{31}{15}\approx2.067\end{aligned}$$
Equivalent decimal values are accepted. The guide also accepts another method, such as finding the percentile for School B, 99.87%.
Marker: determines $z$-score for student in school A
Both heights are normally distributed within their school, so a larger $z$-score means a higher percentile. Write the comparison down.
$$\frac{31}{15}<3$$
This mark may be implied by subsequent working. As percentiles, our own check: $P(Z<2.067)\approx98.06\%$ at School A and $P(Z<3)\approx99.87\%$ at School B.
Marker: provides statement to justify decision
Now name the student. The School B student is further above their school’s mean, measured in standard deviations, so a bigger share of their school is shorter than them.
You never need School B’s mean or standard deviation. The $z$-score already places the student within their own school.
Marker: determines higher ranked student
Each phrase of the question gave you something
Each part is a routine normal-distribution skill: a probability, a quantile, then a $z$-score comparison. Your calculator does parts a) and b) in one command each once the inputs are right.
If I had to guess where you might lose marks, it would be part b). The inverse normal works with the area to the left, so the top 2% needs 0.98. Entering 0.02 gives about 134.194 cm, the bottom 2%. Then round up for a minimum, and draw the curve, because the representation is a mark of its own. In part c), write the comparison $\frac{31}{15}<3$ before you name the student.
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 Mathematical Methods 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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