According to the Bureau of Labor Statistics, 7.1% of the labor force was recently unemployed. A random sample of 100 employable adults was selected. Using the normal approximation to the binomial distribution, what is the probability that 6 or more people from this sample are unemployed?

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Quizplus · Accepted Answer

To solve this, we first find the mean (μ) and standard deviation (σ) of the binomial distribution, then use the normal approximation. For a binomial distribution, μ = np and σ = sqrt(np(1-p)), where n is the sample size and p is the probability of success (in this case, unemployment).Given: n = 100, p = 0.071.μ = np = 100 * 0.071 = 7.1σ = sqrt(np(1-p)) = sqrt(100 * 0.071 * (1 - 0.071)) ≈ sqrt(6.603) ≈ 2.57We're looking for P(X ≥ 6). To use the normal approximation, we apply continuity correction and look for P(X > 5.5) since we're dealing with discrete data.Z = (X - μ) / σ = (5.5 - 7.1) / 2.57 ≈ -0.623Using Z-tables or a calculator, we find the probability to the right of Z = -0.623, which is approximately 0.7324. This means there's a 73.24% chance that 6 or more people from the sample are unemployed.