Final answer:
The probability that the mean weight of the sampled babies differs from the population mean by less than 45 grams, given the large sample size, is practically zero, due to an extreme discrepancy between the sample and population means.
Step-by-step explanation:
The Central Limit Theorem (CLT) suggests that for large sample sizes, the distribution of sample means becomes approximately normally distributed regardless of the original population's distribution. Given the population mean weight of 3366 grams and a population variance of 244,036 grams, we determine the standard deviation of the sample mean, known as the standard error of the mean (SEM), using the formula SEM = σ / √n, where σ is the population standard deviation and n is the sample size. For this case, the SEM is √(244,036 / 118) ≈ 48.822 grams.
To calculate the probability that the sample mean differs by less than 45 grams from the population mean, we convert this difference into a Z-score using the formula Z = (X - μ) / SEM, where X is the difference in grams (45 grams), μ is the population mean (3366 grams), and SEM is the standard error of the mean. Substituting these values, Z = (45 - 3366) / 48.822 ≈ -68.314.
From the standard normal distribution table, we find the probability associated with a Z-score of -68.314. However, Z-scores are not tabulated at such extreme values, implying an incredibly rare event with an extremely low probability. Consequently, the probability that the mean weight of the sample babies differs from the population mean by less than 45 grams is essentially zero due to the vast discrepancy between the sample mean and the population mean.