Question 1
The scores obtained on a test can be assumed to be normally distributed with a mean of 102 and a standard deviation of 19.
What proportion of scores are over 113?
That’s it. Step through the hints below to see why. In 2021, 86.0% of Queensland students chose (A).
Not quite. 0.5789 is the $z$-score of 113, $\tfrac{113-102}{19}\approx0.5789$. It says how many standard deviations 113 is above the mean. It is a distance, not an area, so it isn’t a proportion of scores. In 2021, 7.2% of Queensland students chose (B).
Not quite. 0.7187 is $P(X\lt113)$, the proportion of scores under 113. “Over 113” is the area on the right of 113: $1-0.7187=0.2813$. In 2021, 5.4% of Queensland students chose (C).
Not quite. 113 is above the mean of 102, so fewer than half the scores are over it. Any answer above 0.5, like 0.8216, can’t be right. Check which side of 113 your calculator is finding the area for. In 2021, 1.0% of Queensland students chose (D).
This is the question’s area. About 28.13% of scores are over 113, so the proportion is 0.2813. 113 is above the mean, so the area had to be less than half.
This shades the scores under 113, $P(X\lt113)\approx0.7187$. That is option (C). “Over 113” is the area on the other side of the line.
The cut-off is at the mean, so exactly half the scores are on each side.
About of scores are . The question’s cut-off is 113.
The second row under the axis gives each score as a $z$-score. 113 sits at $z\approx0.58$, a distance along the axis, not an area.
Mark 113 on a normal curve centred at the mean, 102. “Over 113” is the area to the right of 113.
$$P(X\gt113)$$
113 is above the mean, so this area is less than half. That rules out (C) and (D).
Use the normal cdf on your calculator with a lower bound of 113, a very large upper bound, $\mu=102$ and $\sigma=19$. Or standardise first.
$\displaystyle z=\frac{113-102}{19}$$\displaystyle \approx0.5789$
That $z$-score is a distance, not a probability, so (B) isn’t the answer yet.
Find the area to the right of that $z$-score.
$\displaystyle P(Z\gt0.5789)$$\displaystyle \approx0.2813$
It is under 0.5, as the sketch said. The left tail, 0.7187, is the proportion under 113.
$\displaystyle P(X\gt113)$$\displaystyle =P\Bigl(Z\gt\frac{113-102}{19}\Bigr)$
$\displaystyle =P(Z\gt0.5789)$$\displaystyle \approx0.2813$
113 is above the mean, so the proportion of scores over it is less than half. About 28% of scores are over 113.
How Queensland students answered in 2021
From the QCAA Mathematical Methods subject report for 2021. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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