Final answer:
The probability that the average percent of fat calories consumed by a sample of size 16 is more than 30 is approximately 0.9918.
Step-by-step explanation:
In order to find the probability that the average percent of fat calories consumed by a sample of size 16 is more than 30, we can use the Central Limit Theorem. The Central Limit Theorem states that if the sample size is large enough, the sample mean will be approximately normally distributed regardless of the shape of the population distribution.
Since the population distribution is approximately normal with a mean of 36 and a standard deviation of 10, we can calculate the z-score for an average percent of fat calories of 30 using the formula:
z = (x - μ) / (σ / sqrt(n)), where x is the desired average percent of fat calories, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Plugging in the values, we get:
z = (30 - 36) / (10 / sqrt(16)) = -6 / 2.5 = -2.4
Next, we can use a standard normal distribution table or a calculator to find the probability that a z-score is less than -2.4. Looking up the z-score, we find that the probability is approximately 0.0082.
Finally, to find the probability that the average percent of fat calories consumed is more than 30, we subtract the probability of the z-score being less than -2.4 from 1:
Probability = 1 - 0.0082 = 0.9918
Therefore, the probability that the average percent of fat calories consumed by a sample of size 16 is more than 30 is approximately 0.9918. The closest option to this probability from the given choices is D) 0.75.