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June 2013 p63 q4
3320
In a certain country, on average one student in five has blue eyes. For a random selection of 120 students, find the probability that fewer than 33 have blue eyes.
Solution
Let the random variable \(X\) represent the number of students with blue eyes. Given that one in five students has blue eyes, the probability \(p\) is \(0.2\).
The expected number of students with blue eyes is \(\mu = 120 \times 0.2 = 24\).
The variance is \(\sigma^2 = 120 \times 0.2 \times 0.8 = 19.2\).
We approximate \(X\) by a normal distribution \(N(24, 19.2)\).
To find \(P(X < 33)\), we use a continuity correction: \(P(X < 33) \approx P(Z < \frac{32.5 - 24}{\sqrt{19.2}})\).
Calculate the standard score: \(Z = \frac{32.5 - 24}{\sqrt{19.2}} = 1.9398\).
Using standard normal distribution tables, \(P(Z < 1.9398) = 0.974\).
