To assess the impact of undercoverage, nonresponse, and other sampling errors, one can compare sample data with known population figures. In Canada, approximately 24% of the population aged over 15 are first-generation, having been born outside of the country. Consequently, in random samples of 1000 individuals aged over 15, the number of first-generation Canadians (X) should conform to a binomial distribution with n = 1000 and p = 0.24. (a) The mean (μ) and standard deviation (σ) of X are: Mean (μ) = np = 1000 × 0.24 = 240 Standard deviation (σ) = √(np(1 - p)) = √(1000 × 0.24 × 0.76) ≈ 13.8564. (b) To calculate the probability of the sample containing between 210 and 270 first-generation Canadians using normal approximation: First, verify the appropriateness of the normal approximation by ensuring: 1. np ≥ 5 2. n(1 - p) ≥ 5 Since np = 240 and n(1 - p) = 760, both criteria are met. Next, determine the z-scores for 210 and 270: Z for 210 = (210 - μ) / σ = (210 - 240) / 13.8564 ≈ -2.166 Z for 270 = (270 - μ) / σ = (270 - 240) / 13.8564 ≈ 2.166 Then, using a standard normal distribution table or calculator, find the probabilities for these z-scores and calculate the probability that X is between 210 and 270. Probability(X is between 210 and 270) ≈ P(Z < 2.166) - P(Z < -2.166).

Question

To assess the impact of undercoverage, nonresponse, and other sampling errors, one can compare sample data with known population figures. In Canada, approximately 24% of the population aged over 15 are first-generation, having been born outside of the country. Consequently, in random samples of 1000 individuals aged over 15, the number of first-generation Canadians (X) should conform to a binomial distribution with n = 1000 and p = 0.24. (a) The mean (μ) and standard deviation (σ) of X are: Mean (μ) = np = 1000 × 0.24 = 240 Standard deviation (σ) = √(np(1 - p)) = √(1000 × 0.24 × 0.76) ≈ 13.8564. (b) To calculate the probability of the sample containing between 210 and 270 first-generation Canadians using normal approximation: First, verify the appropriateness of the normal approximation by ensuring: 1. np ≥ 5 2. n(1 - p) ≥ 5 Since np = 240 and n(1 - p) = 760, both criteria are met. Next, determine the z-scores for 210 and 270: Z for 210 = (210 - μ) / σ = (210 - 240) / 13.8564 ≈ -2.166 Z for 270 = (270 - μ) / σ = (270 - 240) / 13.8564 ≈ 2.166 Then, using a standard normal distribution table or calculator, find the probabilities for these z-scores and calculate the probability that X is between 210 and 270. Probability(X is between 210 and 270) ≈ P(Z < 2.166) - P(Z < -2.166).

Mathematics · College · Thu Feb 04 2021

llm

Answered on Sat Nov 18 2023

To calculate the mean (μ) and the standard deviation (σ) for the binomial distribution described, you correctly used the formulas:

1000 * 0.24 = 240 σ = √(1000 * 0.24 * (1 - 0.24)) = √(1000 * 0.24 * 0.76) ≈ 13.8564

For the normal approximation, you initiated the process correctly by verifying that the sample distribution satisfies the criteria for a normal approximation:

1. np = 240 ≥ 5 2. n(1 - p) = 760 ≥ 5

Both conditions are met, thus a normal approximation can be used.

Next, you calculated the z-scores correctly:

Z for 210 = (210 - μ) / σ = (210 - 240) / 13.8564 ≈ -2.166 Z for 270 = (270 - μ) / σ = (270 - 240) / 13.8564 ≈ 2.166

Then, you will need to find the area under the normal curve corresponding to these z-scores. Unfortunately, a slight correction is needed here. The probability that X is between 210 and 270 will be found by:

Probability(X is between 210 and 270) ≈ P(Z ≤ 2.166) - P(Z ≤ -2.166)

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