In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? Assume that we want 96% confidence that the error
Is no more than 2.5 percentage points.
A) 1682
B) 591
C) 1281
D) 811

Question

Quizplus · Accepted Answer

The sample size required to estimate the percentage of people who weigh more than 180 pounds with a 96% confidence level and a margin of error of 2.5 percentage points is 1682. This is calculated using the formula: n = (Z^2 * p * q) / e^2, where Z is the z-score for the desired confidence level (2.576 for 96% confidence), p is the estimated proportion of people who weigh more than 180 pounds (0.5), q is the complement of p (0.5), and e is the margin of error (0.025).

In a Certain Population, Body Weights Are Normally Distributed with a Mean