9.3: The Central Limit Theorem for Sample Proportions
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9.3: The Central Limit Theorem for Sample Proportions
Opening the book…
Theorem \(\PageIndex{1}\) Central Limit Theorem for Sample Proportions
If \(p\) is the proportion of a population that possesses a certain feature, then \(\hat{p}\), the proportion of a sufficiently large sample of size \(n\) that possess the same feature, is approximately normally distributed with the mean \(\mu_{\hat{p}}=p\) and the standard deviation \(\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}\). In this context, a "large" sample is such that \(np\geq10\) and \(n(1-p)\geq10\).
Symbolically, the statement can be written in the following way:
For \(X\sim?(p)\), when \(np\geq10\) and \(n(1-p)\geq10\),
According to a study, 5.4% of the passengers fly first-class. A random sample of size 215 was obtained by a researcher. Find the probability that at least 10 people in the sample are first-class passengers.
Solution
Let \(p\) be the proportion of people that fly first-class, then \(p=0.054\).
Let \(\hat{p}\) be the proportion of people that fly first-class in the sample of 215 random passengers. Since the sample is large, i.e.
\(np=215\cdot0.054=11.61\geq10\) and \(n(1-p)=215\cdot(1-0.054)=215\cdot0.946=203.39\geq10\)