The lengths of pregnancies in a small rural village are normally distributed with a mean of 265 days and a standard deviation of 13 days. In what range would you expect turind the middle \( 98 \% \) of most pregnancies? Between If you were to draw samples of size 49 from this population, in what range would you expect to find the middle \( 98 \% \) of most averages for the lengths of pregnancies in the sample? Between

To find the range for the middle 98% of most pregnancies in the population, we start by determining the z-scores that correspond to the middle 98%. This leaves 2% in the tails, which means 1% in each tail. From the standard normal distribution table, the z-scores for 1% in each tail are approximately ±2.33. We can calculate the range for the population as follows: - Lower Range = Mean - (z-score * Standard Deviation) = 265 - (2.33 * 13) ≈ 236.71 days - Upper Range = Mean + (z-score * Standard Deviation) = 265 + (2.33 * 13) ≈ 293.29 days So, we expect 98% of pregnancies to fall between approximately 237 and 293 days. When drawing samples of size 49, we apply the Central Limit Theorem, which tells us that the sampling distribution of the sample mean will be normally distributed with a mean of 265 days and a standard deviation (standard error) of \(\frac{13}{\sqrt{49}} = \frac{13}{7} \approx 1.86\) days. For the middle 98% of sample means, we use the same z-scores (±2.33): - Lower Range = Mean - (z-score * Standard Error) = 265 - (2.33 * 1.86) ≈ 261.68 days - Upper Range = Mean + (z-score * Standard Error) = 265 + (2.33 * 1.86) ≈ 268.32 days Therefore, we would expect the middle 98% of averages from samples of size 49 to fall between approximately 261.68 and 268.32 days.