Given a population where the proportion of items with a desired attribute is p = 0.25, if a sample of 400 is taken, what is the probability the proportion of successes in the sample will be greater than 0.22?
A) 0.9162
B) 0.8812
C) 0.7141
D) 0.8412

Question

Quizplus · Accepted Answer

This problem can be solved using the normal approximation to the binomial distribution. The mean of the sampling distribution of the proportion is $p = 0.25$, and the standard deviation is $\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.25(0.75)}{400}} = 0.02165$. We want to find the probability that the sample proportion is greater than 0.22, so we calculate the z-score for 0.22: $z = \frac{0.22 - 0.25}{0.02165} = -1.387$. Looking up this z-score in a standard normal distribution table or using a calculator gives us the probability of being less than 0.22. Since we want the probability of being greater, we subtract this value from 1. The exact probability might vary slightly depending on the method of calculation, but the closest answer provided is 0.9162, indicating a high likelihood of the sample proportion being greater than 0.22.

Given a Population Where the Proportion of Items with a Desired