Respuesta :
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,[tex]\bar X\sim N(\mu,\ \frac{\sigma^{2}}{{n}})[/tex]
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:
[tex]CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}[/tex]
[tex]=267\pm 2.33\cdot\frac{17}{\sqrt{47}}\\\\=267\pm5.78\\\\=(261.22, 272.78)\\\\\approx (261, 273)[/tex]
Thus, the range of middle 98% of most averages for the lengths of pregnancies in the sample is, (261, 273).
