Final answer:
The sampling distribution of the proportion of subjects with lung cancer in a sample of 5,000 non-smokers can be modeled using the binomial distribution. The mean of the proportion is 4 and the standard deviation is 0.6364.
Step-by-step explanation:
The sampling distribution of the proportion of subjects with lung cancer in a sample of 5,000 non-smokers can be modeled using the binomial distribution. The binomial distribution is appropriate when we have a fixed number of trials (in this case, 5,000) and each trial can result in one of two outcomes (having lung cancer or not having lung cancer). The proportion of non-smokers with lung cancer in the population is given as 80 per 100,000, which is equivalent to 0.0008.
To find the sampling distribution of the proportion, we can calculate the mean and standard deviation of the binomial distribution. The mean (μ) is equal to the product of the sample size (n) and the probability of success (p), i.e., μ = n * p. In this case, μ = 5,000 * 0.0008 = 4. The standard deviation (σ) is equal to the square root of the product of the sample size, the probability of success, and the probability of failure (1-p), i.e., σ = sqrt(n * p * (1-p)). Plugging in the values, we get σ = sqrt(5,000 * 0.0008 * (1 - 0.0008)) ≈ 0.6364.
Therefore, the sampling distribution of the proportion of subjects with lung cancer in the sample of 5,000 non-smokers is approximately normal, with a mean of 4 and a standard deviation of 0.6364.