On average, 34% of the people who go to a particular theatre are men.
Use an approximation to find the probability that, in a random sample of 600 people who go to the theatre, fewer than 190 are men.
Solution
Let the random variable representing the number of men be denoted by a binomial distribution, \(X \sim B(n, p)\), where \(n = 600\) and \(p = 0.34\).
The mean \(\mu\) is given by \(np = 600 \times 0.34 = 204\).
The variance \(\sigma^2\) is given by \(np(1-p) = 600 \times 0.34 \times 0.66 = 134.64\).
We approximate the binomial distribution with a normal distribution \(N(\mu, \sigma^2)\).
To find \(P(X < 190)\), we use a continuity correction: \(P(X < 190) \approx P(X < 189.5)\).
Standardize using \(Z = \frac{X - \mu}{\sigma}\):
\(P(X < 189.5) = P\left( Z < \frac{189.5 - 204}{\sqrt{134.64}} \right) = P(Z < -1.2496)\).
Using the standard normal distribution table, \(P(Z < -1.2496) = 1 - \Phi(1.2496)\).
\(\Phi(1.2496) \approx 0.8944\), so \(1 - 0.8944 = 0.106\).
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