Question 15
A tour operator offers day cruises off the Queensland coast. They advertise that on any given day, customers have a 65% probability of seeing at least one whale.
The tour operator conducts cruises for 40 consecutive days.
Let $X$ be the binomial random variable representing the outcome — success or failure — over the 40 days. A success is customers seeing at least one whale.
Calculate the probability that the number of days with a success will be within one standard deviation of the mean for $X$.
[3 marks]Forty cruises, one count, a band around the mean
Each day is a success or a failure. Count the successes, then see which counts sit within one standard deviation of the mean.
The band runs from 22.98 to 29.02, but $X$ only takes whole numbers. The counts inside are 23 to 29.
Turn the band into whole-number bounds
The band $\mu\pm\sigma$ has decimal ends, but your calculator’s binomial cumulative function needs whole-number bounds. Set the lower and upper bounds so the blue bars are exactly the counts inside the band.
9.1 is the variance, $np(1-p)$. The standard deviation is its square root, about 3.0166. A band of 9.1 either side covers almost every possible count.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The mean and standard deviation are one mark each. The last mark is the probability, which needs the right whole-number bounds.
Each day is either a success or a failure, so $X$ is binomial. Read off the number of trials and the probability of success, then use the binomial mean.
$$\begin{aligned}p&=0.65\\n&=40\\\mu&=np\\&=40\times0.65\\&=26\end{aligned}$$
Marker: correctly determines the mean
The standard deviation is the square root of the variance, $np(1-p)$.
$$\begin{aligned}\sigma&=\sqrt{40\times0.65\times0.35}\\&=\sqrt{9.1}\\&\approx3.0166\end{aligned}$$
Marker: correctly determines the standard deviation
One standard deviation either side of the mean runs from about 22.98 to 29.02. $X$ counts days, so it only takes whole numbers. The counts inside the band are 23 to 29.
$$\begin{aligned}\text{Lower: }26-3.0166&\approx22.98\ \Rightarrow\ 23\\\text{Upper: }26+3.0166&\approx29.02\ \Rightarrow\ 29\end{aligned}$$
The sample response writes these as $26-3=23$ and $26+3=29$. Use your calculator’s binomial cumulative function with $n=40$, $p=0.65$, a lower bound of 23 and an upper bound of 29.
$$P(23\le X\le29)\approx0.7547$$
The sample response gives the answer as 0.7547 and as 75%.
Marker: determines the required probability
Each phrase of the question gave you something
Three routines are chained together. You find the mean and the standard deviation, turn them into a band with decimal ends, then turn that band into whole-number bounds before the calculator can do anything. The step in the middle is the one you have to spot for yourself, because the question never says which counts to include. It stays complex familiar because the question names the binomial model and spells out “within one standard deviation of the mean”, so the method is clear from the wording.
If I had to guess where you might lose marks, it would be the bounds. Rounding outwards to 22 and 30 gives about 0.8656, and leaving out the end values 23 and 29 gives about 0.5925. Both look like sensible probabilities, so nothing warns you. The other trap is stopping at the variance. 9.1 is $np(1-p)$, and the standard deviation is its square root.
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 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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