Question

the lengths of pregnancies in a small rural village are normally distributed with a mean of 264 days and a standard deviation of 16 days.

in what range would you expect to find the middle 68% of most pregnancies?
between days and days.

if you were to draw samples of size 33 from this population, in what range would you expect to find the middle 68% of most averages for the lengths of pregnancies in the sample?
between days and days.

enter your answers as numbers to the nearest whole days.

Explanation:

Identify the population parameters

We are given that the lengths of pregnancies are normally distributed with:

• Population mean \(\mu = 264\) days
• Population standard deviation \(\sigma = 16\) days

Calculate the range for individual pregnancies

Using the Empirical Rule (68-95-99.7 rule), the middle \(68\%\) of a normal distribution lies within \(1\) standard deviation of the mean:

$$ \text{Range} = \mu \pm 1\sigma $$

$$ \text{Lower limit} = 264 - 16 = 248 $$

$$ \text{Upper limit} = 264 + 16 = 280 $$

Calculate the standard error for sample means

Using the Sampling Distribution of the Mean and Standard Error of the Mean concepts, for sample size \(n = 33\):

$$ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{33}} \approx 2.785 \text{ days} $$

Calculate the range for sample averages

The middle \(68\%\) of sample means lies within \(1\) standard error of the population mean:

$$ \text{Range} = \mu \pm 1\sigma_{\bar{x}} $$

$$ \text{Lower limit} = 264 - 2.785 = 261.215 \approx 261 $$

$$ \text{Upper limit} = 264 + 2.785 = 266.785 \approx 267 $$

Answer:

Question 1

In what range would you expect to find the middle 68% of most pregnancies?
Between <blank>248</blank> days and <blank>280</blank> days.

Question 2

If you were to draw samples of size 33 from this population, in what range would you expect to find the middle 68% of most averages for the lengths of pregnancies in the sample?
Between <blank>261</blank> days and <blank>267</blank> days.