A simple random sample of size n = 1460 is obtained from a population whose size is N = 1,500,000 and whose population proportion with a specified characteristic is p = 0.42.


c. What is the probability of obtaining x = 584 or fewer individuals with the characteristic?

Verified step by step guidance

1. Identify the parameters of the problem: population size \(N = 1,500,000\), sample size \(n = 1460\), population proportion \(p = 0.42\), and the value of interest \(x = 584\) individuals with the characteristic.
2. Since the sample size is large, approximate the distribution of the number of individuals with the characteristic, \(X\), using a normal distribution. First, calculate the mean \(\mu\) and variance \(\sigma^2\) of \(X\) using the formulas for a hypergeometric distribution or, more commonly, the binomial approximation with finite population correction:
3. \[\mu = n \times p\]
4. \[\sigma^2 = n \times p \times (1 - p) \times \frac{N - n}{N - 1}\]
5. Calculate the standard deviation \(\sigma\) by taking the square root of the variance:
6. \[\sigma = \sqrt{\sigma^2}\]
7. Convert the discrete value \(x = 584\) to a continuous value for the normal approximation by applying the continuity correction. Use \(x + 0.5\) or \(x - 0.5\) depending on the direction of the inequality. Here, since we want \(P(X \leq 584)\), use \(x = 584 + 0.5 = 584.5\).
8. Standardize the value to find the corresponding \(z\)-score in the normal distribution:
9. \[z = \frac{584.5 - \mu}{\sigma}\]
10. Finally, use the standard normal distribution table or a calculator to find the probability \(P(Z \leq z)\), which approximates \(P(X \leq 584)\).

Key Concepts

Sampling Distribution of the Sample Proportion

The sampling distribution of the sample proportion describes how the proportion of individuals with a characteristic varies across different samples of the same size. It is approximately normal for large samples, with mean equal to the population proportion p and standard deviation depending on p and the sample size n.

Normal Approximation to the Binomial Distribution

When the sample size is large, the binomial distribution of the number of successes can be approximated by a normal distribution. This simplifies probability calculations, using the mean np and standard deviation sqrt(np(1-p)), especially when np and n(1-p) are both greater than 5.

Continuity Correction

The continuity correction adjusts for the fact that the binomial distribution is discrete while the normal distribution is continuous. When approximating P(X ≤ x) using the normal distribution, we use P(X ≤ x + 0.5) to improve accuracy in probability estimates.