Final answer:
To determine the probability of observing a sample mean of 21 or more from a sample of size 36, use the Central Limit Theorem and calculate the z-score. The probability is approximately 3.59%.
Step-by-step explanation:
To determine the probability of observing a sample mean of 21 or more from a sample of size 36, we need to use the sampling distribution of the sample mean. Since the population is left skewed and the sample size is large (n >= 30), we can use the Central Limit Theorem.
The Central Limit Theorem states that the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is large enough. The mean of the sampling distribution will be equal to the population mean, which is 24.
To find the probability of observing a sample mean of 21 or more, we need to calculate the z-score using the formula:
z = (sample mean - population mean) / (standard deviation / sqrt(sample size))
In this case, the sample mean is 21, population mean is 24, standard deviation is 10, and sample size is 36. Plugging in these values, we get:
z = (21 - 24) / (10 / sqrt(36))
Calculating the z-score gives us:
z = -1.8
Next, we need to find the area under the standard normal curve to the right of the z-score of -1.8. We can use a standard normal table or a calculator to find this area, which represents the probability.
Using a standard normal table, the area to the right of -1.8 is approximately 0.9641. Therefore, the probability of observing a sample mean of 21 or more from a sample of size 36 is approximately 0.0359, or 3.59%.