Answer:
To answer these, we can use the standard normal distribution & z-scores. A z-score represents the number of standard deviations a value is from the mean. The formula to calculate a z-score is: z = (x - μ) / σ, where x is the value, μ is the mean and σ is the standard deviation.
A) To find the proportion of pregnancies that last more than 278 days, we first calculate the z-score for 278 days: z = (278 - 272) / 12 = 0.5. Using a standard normal distribution table, we find that the proportion of values above a z-score of 0.5 is approximately 0.3085. So, about 30.85% of pregnancies last more than 278 days.
B) To find the proportion of pregnancies that last between 257 and 275 days, we first calculate the z-scores for both values: z1 = (257 - 272) / 12 = -1.25 and z2 = (275 - 272) / 12 = 0.25. Using a standard normal distribution table, we find that the proportion of values between z-scores of -1.25 and 0.25 is approximately 0.3944. So, about 39.44% of pregnancies last between 257 and 275 days.
C) To find the probability that a randomly selected pregnancy lasts no more than 266 days, we first calculate the z-score for 266 days: z = (266 - 272) / 12 = -0.5. Using a standard normal distribution table, we find that the proportion of values below a z-score of -0.5 is approximately 0.3085. So, there is about a 30.85% chance that a randomly selected pregnancy lasts no more than 266 days.
D) To determine if very preterm babies are unusual, we first calculate the z-score for 242 days: z = (242 - 272) / 12 = -2.5. Using a standard normal distribution table, we find that the proportion of values below a z-score of -2.5 is approximately 0.0062. Since this value is less than 0.05, we can conclude that very preterm babies are unusual.