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1989 Paper 3 Q16
D: 1700.0 B: 1484.0
probability conditional probability approximating binomial to normal distribution normal approximation hypothesis testing population model joint probability table continuity correction

It is believed that the population of Ruritania can be described as follows:

  1. \(25\%\) are fair-haired and the rest are dark-haired;
  2. \(20\%\) are green-eyed and the rest hazel-eyed;
  3. the population can also be divided into narrow-headed and broad-headed;
  4. no narrow-headed person has green eyes and fair hair;
  5. those who are green-eyed are as likely to be narrow-headed as broad-headed;
  6. those who are green-eyed and broad-headed are as likely to be fair-headed as dark-haired;
  7. half of the population is broad-headed and dark-haired;
  8. a hazel-eyed person is as likely to be fair-haired and broad-headed as dark-haired and narrow-headed.
Find the proportion believed to be narrow-headed. I am acquainted with only six Ruritanians, all of whom are broad-headed. Comment on this observation as evidence for or against the given model. A random sample of 200 Ruritanians is taken and is found to contain 50 narrow-heads. On the basis of the given model, calculate (to a reasonable approximation) the probability of getting 50 or fewer narrow-heads. Comment on the result.

Solution:

TikZ diagram
Conditions tell us: \begin{align*} && a+b+d+e &= 0.25 \\ && b+c+e+f &= 0.2 \\ && e &= 0 \\ && b+c &= e + f \\ && b &= c \\ && c+h &= 0.5 \\ && a &= g \\ \end{align*}
TikZ diagram
So \(4b = 0.2 \Rightarrow b = 0.05\)
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And \begin{align*} && 0.25 &= a + d + 0.05 \\ && 1 &= 2a + d + 0.65 \\ \Rightarrow && a &= 0.15 \\ && d &= 0.05 \end{align*}
TikZ diagram
So the proportion who are narrow-headed is \(30\%\). It's obviously relatively unlikely for your six Ruritanian friends to all be broad-headed if it's a random sample, but friendship groups are are likely to be biased so it's not too surprising. Assuming there is a sufficiently large number of Ruritanians, we might model the number of narrow-headed Ruritanians from a sample of \(200\) as \(X \sim B(200, 0.3)\). Computing \(\mathbb{P}(X \leq 50)\) by hand is tricky, so let's use a binomial approximation to obtain: \(X \approx N(60, 42)\) and \begin{align*} \mathbb{P}(X \leq 50) &\approx \mathbb{P} \left (Z \leq \frac{50 - 60+0.5}{\sqrt{42}} \right) \\ &\approx \mathbb{P} \left (Z \leq -\frac{9.5}{6.5} \right) \\ &\approx \mathbb{P} \left (Z \leq -\frac{3}{2} \right) \\ &\approx 5\% \end{align*} (actually this approximation gives \(7.1\%\) and the binomial value gives \(7.0\%\)). This also seems somewhat surprising

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