Question

Mean and Standard Deviation of a Proportion To find the proportion of times something occurs, we divide the count (often a binomial random variable) by the number of trials \(n\). Using the formula for the mean and standard deviation of a binomial random variable, derive the mean and standard deviation of a proportion resulting from \(n\) trials and probability of success \(p\).

Short answer

The mean and standard deviation of a proportion resulting from \(n\) trials and probability of success \(p\) are, respectively, \(p\) and \(\sqrt{\frac{pq}{n}}\).

Step-by-step solution

Understanding Mean and Standard Deviation

For a binomial random variable \(X\), the mean \(\mu\) and standard deviation \(\sigma\) are given by \(\mu=np\), and \(\sigma=\sqrt{npq}\), respectively, where \(q=1-p\). A proportion is derived by dividing the count \(X\) by the number of trials \(n\). Therefore, if \(P\) is the proportion, then we can write \(P=\frac{X}{n}\).

Finding the Mean of the Proportion

To find the mean of the proportion, use the equation for the mean of \(X\) but substitute \(X\) with \(nP\). This gives: \(nP = np\). By dividing throughout by \(n\), the result is: \(P = p\). Thus, the mean of the proportion is equal to the probability of success \(p\).

Finding the Standard Deviation of the Proportion

The standard deviation of the proportion is derived by substituting \(X\) with \(nP\) in the formula for the standard deviation of \(X\). This gives: \(n\sigma = \sqrt{npq}\). We divide by \(n\) to isolate \(\sigma\), resulting in \(\sigma = \sqrt{\frac{pq}{n}}\). Hence, the standard deviation of the proportion is \(\sqrt{\frac{pq}{n}}\).