Final answer:
To calculate the probability that the mean childhood asthma prevalence for a sample of 33 cities is greater than 2.7%, we can use the Central Limit Theorem and standardize the sample mean. The probability obtained represents the likelihood of the sample mean being greater than 2.7%.
Step-by-step explanation:
To answer this question, we can use the Central Limit Theorem. According to the information given, the mean percent of childhood asthma prevalence in the population of 43 cities is 2.29% with a standard deviation of 1.30%. Since the number of cities is large enough (43 cities > 30), we can assume that the sampling distribution of the sample mean will be approximately normally distributed.
To calculate the probability that the mean childhood asthma prevalence for a sample of 33 cities is greater than 2.7%, we first need to standardize the sample mean using the formula: (sample mean - population mean) / (population standard deviation / sqrt(sample size)). Plugging in the values, we get: (2.7 - 2.29) / (1.30 / sqrt(33)).
Next, we need to find the z-score associated with this standardized sample mean. We can use a z-table or a calculator to find the probability of getting a z-score greater than the one we calculated. This probability represents the probability that the mean childhood asthma prevalence for the sample is greater than 2.7%.
The interpretation of this probability is that there is a certain likelihood that the mean childhood asthma prevalence for a random sample of 33 cities is greater than 2.7%. The higher the probability, the more likely it is that the sample mean will be greater than 2.7%.