2022 Paper 2, Q18
The intelligence quotient (IQ) of individuals in a population is normally distributed, with a mean of 100 and a standard deviation of 16.
Nine individuals are chosen at random from the population.
Determine the probability that no more than two of the individuals have an IQ of at least 120.
One distribution, then nine trials
Each person chosen is a trial with the same 10.57% chance of scoring above 120. Watch four sample groups. Most of them pass the “no more than two” test, but not all of them.
The normal answer is only the ingredient
120 is $\frac{120-100}{16}=1.25$ standard deviations above the mean, so $P(IQ\ge 120)=0.1057$. For a continuous distribution “at least 120” and “above 120” are the same thing.
“Nine individuals are chosen at random” turns one probability into nine independent trials, each a success or a failure. That recognition is a mark of its own.
$P(X\le 2)=P(0)+P(1)+P(2)$ with $X\sim B(9,\,0.1057)$, a cumulative binomial, not a single term.
Nine trials, and we want three of the ten possible outcomes
Drag the cut-off. Almost all of the probability sits in the first three bars. Groups of nine rarely contain three high scorers, even though the tail is one in ten.
Which distribution answers the question?
QCAA marking guide · 3 marks
$$P(IQ\ge 120)=0.1057$$
Marker: correctly determines the probability of $IQ\ge 120$. Equivalent decimal values accepted. Students are not required to provide a diagram.
Using a binomial distribution with $n=9$ and $p=0.1057$.
Marker: correctly recognises the context is suitable for modelling as a binomial. Bernoulli trials or a Bernoulli random variable are also accepted. This is the mark that is lost by going straight to the calculator.
$$P(X\le 2)=0.9391$$
About a 94% chance
Marker: determines the required probability. FT mark allowed for an incorrect $p$, and values depending on earlier rounding are accepted, so carrying 0.1057 or 0.10565 both score.
The answer is 0.9391, made from three terms
$P(0)=0.3662$, $P(1)=0.3892$, $P(2)=0.1839$. They sum to 0.9393. The tiny gap from 0.9391 is the rounding of $p$.
$1-P(X\ge 3)$ works too, but you have to add up the terms the other way. With $np=0.95$, three or more high scorers in a group of nine is unusual, and it only happens in about 6% of groups.
$z=1.25$ gives 0.1057 here, and the 2024 e-scooter question uses the identical tail (23 km/h against a mean of 18 with $\sigma=4$). It is worth recognising on sight.
This is three marks, and one of them is for a sentence. The words “binomial”, “trial” and “independent” never appear. What does appear is “nine individuals are chosen at random”, and the marker is looking for you to notice that this is nine repetitions of the same event. Write that sentence down! The other trap is easier to miss. “No more than two” sounds like a single value, so I see students calculate $P(X=2)$ and stop. It is a cumulative probability over three outcomes, and it is the only question in the whole set where you use two distributions in the same answer.
Question wording and marking-guide steps are from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.