Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 264 days and a standard deviation of 17 days.In what range would you expect to find the middle 95% of most pregnancies?Between and .If you were to draw samples of size 31 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 and .

Answer 1

To solve for the middle range of mean:

The middle 95% of the pregnancies are within two standard deviations of the mean:

The middle 95% of the pregnancies are within two standard deviations of the mean are 230 and 298

The middle 95% of the average length of pregnancies' would be a 95% CI: mean +/- z-critical * standard deviation/square root(sample size)

Critical value for 95% confidence interval = 1.96

`\begin{gathered} 264\pm1.96\cdot\frac{17}{\sqrt[]{31}} \\ =264+\frac{1.96(17)}{\sqrt[]{31}} \\ =264\pm(33.32)/(5.57) \\ =264\pm5.982 \\ =269.982\text{ , 258.018} \end{gathered}`

The middle 95% of the average length of pregnancies' would be a 95% CI: between 258.018 and 269.982