Based on the caterer's experience, 38% of attendees to events will prefer chicken for the main dish.As the caterer plans for an event attended by 780 individuals, the normal approximation will be used for the binomial with a correction for continuity.In this case, what is the probability that less than 300 will prefer chicken?
A)0.604
B)0.380
C)0.590
D)0.620
E)0.618

Question

Quizplus · Accepted Answer

The normal approximation to the binomial distribution is used here. The mean (μ) is np = 780 * 0.38 = 296.4, and the standard deviation (σ) is √(np(1-p)) = √(780 * 0.38 * 0.62) ≈ 13.67. For P(X < 300), use the continuity correction: P(X < 299.5). Convert to a z-score: z = (299.5 - 296.4) / 13.67 ≈ 0.227. Using standard normal distribution tables or a calculator, P(Z < 0.227) ≈ 0.590.

Based on the Caterer's Experience, 38% of Attendees to Events