Question

If you were to draw samples of size 47 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?

Answer 1

Answer:

The range of middle 98% of most averages for the lengths of pregnancies in the sample is, (261, 273).

Step-by-step explanation:

The complete question is:

The lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 17 days. If you were to draw samples of size 47 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?

Solution:

As the sample size is large, i.e. n = 47 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean by the normal distribution.

So,$\bar X\sim N(\mu,\ \frac{\sigma^{2}}{{n}})$

The range of the middle 98% of most averages for the lengths of pregnancies in the sample is the 98% confidence interval.

The critical value of z for 98% confidence level is,

z = 2.33

Compute the 98% confidence interval as follows:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$

     $=267\pm 2.33\cdot\frac{17}{\sqrt{47}}\\\\=267\pm5.78\\\\=(261.22, 272.78)\\\\\approx (261, 273)$

Thus, the range of middle 98% of most averages for the lengths of pregnancies in the sample is, (261, 273).