Question 5
The mass (g) of adult kookaburras in a certain region is normally distributed with a mean of 300 g and a standard deviation of 13 g. Select the correct statement about the mass of adult kookaburras.
Not quite. 287 g to 313 g is $300\pm13$, one standard deviation either side of the mean. That band holds about 68%. 34% is only one half of it, from 300 g to 313 g. In 2024, 7.4% of Queensland students chose (A).
Not quite. 274 g to 326 g is $300\pm2(13)$, two standard deviations either side of the mean. That holds about 95%, not 68%. In 2024, 7.5% of Queensland students chose (B).
Not quite. 261 g is three standard deviations below the mean, but 326 g is only two above. The interval isn’t symmetric, so it holds about 97.6%, not 95%. In 2024, 6.7% of Queensland students chose (C).
That’s it. Step through the hints below to see why. In 2024, 77.6% of Queensland students chose (D).
This is the interval in (A), one standard deviation either side of the mean. It holds about 68%, not 34%. Only the half from 300 g to 313 g holds 34%.
This is the interval in (B), two standard deviations either side of the mean. It holds about 95%, not 68%.
This is the interval in (C). It runs three standard deviations below the mean but only two above, so it holds about 97.6%, not 95%.
This is the interval in (D), three standard deviations either side of the mean. It holds 99.7%, so (D) is the correct statement.
The shaded area is the proportion of adult kookaburras between the two bounds. Each step of 13 g is one standard deviation.
Every mass in the options is a whole number of 13 g steps from the mean, 300 g.
$287=300-13$$274=300-2(13)$$261=300-3(13)$
In the same way, 313 g, 326 g and 339 g are one, two and three standard deviations above the mean.
For any normal distribution:
about 68% lie within one standard deviation of the mean, $\mu\pm\sigma$;
about 95% lie within two, $\mu\pm2\sigma$;
about 99.7% lie within three, $\mu\pm3\sigma$.
(A) is $\mu\pm\sigma$: 68%, not 34%.
(B) is $\mu\pm2\sigma$: 95%, not 68%.
(C) runs from $\mu-3\sigma$ to $\mu+2\sigma$, which isn’t one of the symmetric bands.
(D) is $\mu\pm3\sigma$: 99.7%. That one matches.
$300-3(13)=261$$300+3(13)=339$
So 261 g to 339 g is three standard deviations either side of the mean. About 99.7% of a normal distribution lies in that band.
How Queensland students answered in 2024
From the QCAA Mathematical Methods subject report for 2024. 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 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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