Final answer:
The question aims to calculate the probability that 60 or more out of 100 people don't floss enough using a binomial distribution. The normal approximation could be used, but the provided answers are inconclusive as they do not align with the problem statement. More information or clarification might be necessary to provide a definitive answer.
Step-by-step explanation:
To determine the probability that 60 or more people don't floss enough from a sample of 100 people when the estimate is 65%, the binomial distribution can be used. However, in the provided equation, np(1-p) = 22.75, we can recognize that this is related to the calculation of the variance of a binomial distribution, npq (where q = 1 - p). This implies that the normal approximation could be used because the sample size is large and npq is sufficiently large (greater than 10).
The mean of the distribution would be np, and the standard deviation would be sqrt(npq). We would then use z-scores to find the probability of observing a sample proportion at or above 60%. However, the options given in the question do not seem to directly relate to the parameters of the problem as stated. Specifically, option c is not a probability (as probabilities are between 0 and 1), and there is not enough information provided in the question to calculate using the standard normal distribution.
Therefore, it appears there may be an issue with the data provided in the question. Typically, this probability would be solved using a binomial or normal approximation to the binomial, but given the information at hand, we are unable to confirm one of the provided answers.