Final answer:
The question asks for the probability that more than 60% of a sample of primary care doctors think their patients receive unnecessary medical tests, given a known population proportion. The solution involves calculating the mean, standard deviation for the sample proportion, finding the z-score for 60%, and using the z-score to find the probability using the standard normal distribution.
Step-by-step explanation:
The question involves calculating the probability that more than 60% of a sample of 150 primary care doctors think their patients receive unnecessary medical tests, given that 42 percent of the population of primary care doctors believe this way. This type of problem is typically addressed using the binomial or normal approximation to the binomial distribution. Since the sample size is relatively large, the normal approximation is more appropriate.
To solve this, we first determine the mean (μ) and standard deviation (σ) for the proportion in the sample. The mean is simply n * p, where n is the sample size (150 doctors) and p is the population proportion (0.42). The standard deviation for the proportion is given by √[np(1-p)]. After finding these values, we need to find the z-score for 60% and use standard normal distribution tables, calculators, or technology to find the probability of observing a proportion greater than this z-score.
Steps to Calculate the Probability
- Calculate the mean (μ = 150 * 0.42).
- Calculate the standard deviation (σ = √[150 * 0.42 * (1 - 0.42)]).
- Find the z-score for the sample proportion of 60%.
- Use the z-score to determine the corresponding probability.