Natalia estimates that she wins a board game 72% of the time. How large of a random sample is required to obtain a margin of error of at most 0.06 with 95% confidence?Find the z-table here.

For a proportion, use n = z²p(1 − p) / E². With 95% confidence, z = 1.96; here p = 0.72 and the desired margin of error is E = 0.06. Substituting gives n = (1.96)²(0.72)(0.28) / (0.06)² ≈ 215.13. Round up to ensure the margin of error is at most 0.06, so the required sample size is 216, choice B.

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