Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.At Meadowbrook Hospital,the mean weight of babies born to full-term pregnancies is 7 pounds with a standard deviation of 14 ounces (1 pound = 16 ounces) . Dr.Watts (who works at Meadowbrook Hospital) has four deliveries (all for full-term pregnancies) coming up during the night.Assume that the birth weights of these four babies can be viewed as a simple random sample.What is the probability that all four babies will weigh more than 7.5 pounds?

A) 0.0065
B) 0.1265
C) 0.2839
D) 0.4858

We need to first standardize the birth weight using the formula: z = (x - μ) / σ , where x is the birth weight, μ is the mean birth weight (in this case, 7 pounds = 112 ounces), and σ is the standard deviation (14 ounces). Therefore, z = (120 - 112) / 14 = 0.571. We then use a Normal distribution table (or calculator) to find the probability that a single baby weighs more than 7.5 pounds (i.e., a z-score of 0.3125). This probability is approximately 0.3849. Since each delivery is assumed to be independent, the probability that all four babies weigh more than 7.5 pounds is obtained by multiplying the probabilities together: P(all four > 7.5 pounds) = 0.3849^4 = 0.0065 (rounded to four decimal places). Therefore, the answer is A.