According to a hospital administrator,historical records over the past 10 years have shown that 20% of the major surgery patients are dissatisfied with after-surgery care in the hospital.A scientific poll based on 400 hospital patients has just been conducted. What is the probability that less than 64 patients will not be satisfied with the after-surgery care?
A)47.72%
B)2.28%
C)97.72%
D)95.44%
E)4.56%

Question

Quizplus · Accepted Answer

The probability of a patient being dissatisfied with after-surgery care is 0.20. We are asked to find the probability that less than 64 patients will not be satisfied with the after-surgery care, which is the same as finding the probability that 64 or more patients will be satisfied.
The sample size n = 400 and the probability of dissatisfaction p = 0.20.
We can use the normal approximation to the binomial distribution, with mean np = 80 and standard deviation sqrt(np(1-p)) = 7.97.
To find the probability that 64 or more patients will be satisfied, we can calculate the z-score:
z = (x - np) / sqrt(np(1-p))
where x = 336 (number of satisfied patients in the sample).
z = (336 - 80) / 7.97 = 50.68
Using the normal distribution table, the probability of getting a z-score of 50.68 or higher is very small, approximately 0.0228.
Therefore, the probability that less than 64 patients will not be satisfied with the after-surgery care is 0.0228 or 2.28%. So, the correct answer is B.

According to a Hospital Administrator,historical Records Over the Past 10