9.3: The Central Limit Theorem for Sample Proportions

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    9.3: The Central Limit Theorem for Sample Proportions
    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 \(n(1-p)\geq10\),

    \(\hat{p}\sim N\left(\mu_{\hat{p}}=p, \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}\right)\)

    Try It Yourself! \(\PageIndex{0.1}\)
    Try It Yourself! \(\PageIndex{0.2}\)
    Example \(\PageIndex{1}\)

    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.

    Image of passengers waiting at the airport.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 \(n(1-p)=215\cdot(1-0.054)=215\cdot0.946=203.39\geq10\)

    by CLT,

    \(\hat{p}\sim N\left(\mu_{\hat{p}}=p=0.054, \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.054(1-0.054)}{215}}=0.0154\right)\)

    Then the probability that at least 10 people (i.e., proportion is at least 10/215) in the sample are first-class passengers can be expressed as

    \(P(\hat{p}>\frac{10}{215})=P(\hat{p}>0.0465)=P\left(\dfrac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}>\dfrac{0.0465-0.054}{0.0154}\right)=P(Z>-0.49)=0.6879=68.79\%\)

    Or using technology,

    \(P(\hat{p}>0.0465)=0.6869=68.69\%\)

    Try It Yourself! \(\PageIndex{1}\)
    Try It Yourself! \(\PageIndex{2}\)
    Try It Yourself! \(\PageIndex{3}\)
    Try It Yourself! \(\PageIndex{4}\)
    Try It Yourself! \(\PageIndex{5}\)

    This page titled 9.3: The Central Limit Theorem for Sample Proportions was last modified on Sat, 16 Aug 2025 20:22:30 GMT and is shared under a not declared license and was authored, remixed, and/or curated by Edward Pineda-Castro.