The lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of 13 days. In what range would you expect to find the middle \( 95 \% \) of most pregnancies? Between If you were to draw samples of size 56 from this population, in what range would you expect to find the middle \( 95 \% \) of most averages for the lengths of pregnancies in the sample? Between Enter your answers as numbers. Your answers should be accurate to 1 decimal places.

To find the range for the middle 95% of the pregnancies, we can use the properties of the normal distribution. The range is calculated using the mean and standard deviation. 1. For the population: Mean = \( 262 \) days Standard deviation = \( 13 \) days The z-scores corresponding to the middle 95% are approximately -1.96 and 1.96. Therefore, the range can be calculated as: \[ 262 - (1.96 \times 13) \] to \[ 262 + (1.96 \times 13) \] Which gives: \[ 262 - 25.48 \approx 236.5 \] days to \[ 262 + 25.48 \approx 287.5 \] days. So for the population, you would expect the middle 95% of pregnancies to be between **236.5** and **287.5** days. 2. For the sample of size 56: The standard error (SE) is calculated as the standard deviation divided by the square root of the sample size: \[ SE = \frac{13}{\sqrt{56}} \approx 1.74 \text{ days} \] Using the mean and the z-scores corresponding to the middle 95%: \[ 262 - (1.96 \times 1.74) \] to \[ 262 + (1.96 \times 1.74) \] Which gives: \[ 262 - 3.41 \approx 258.6 \] days to \[ 262 + 3.41 \approx 265.4 \] days. So for the sample averages, you would expect the middle 95% of averages to be between **258.6** and **265.4** days.