Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days.
In what range would you expect to find the middle 98% of most pregnancies?


Between ___ and ___ .



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?


Between ___ and ____ .

Answer 1

Answer:

(221.39, 300.61) and (255.2223, 266.7777)

Step-by-step explanation:

Given that X, the  lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days

Middle 98% would lie on either side of the mean with probability ±2.33 in the std normal distribution on either side of 0

Corresponding we have x scores as

Between $261-2.33*17$ and $261+2.33*17$

i.e. in the interval = (221.39, 300.61)

If sample size = 47, then std error of sample would be

$\frac{17}{\sqrt{47} }$

So 98% of pregnancies would lie between

$261-2.33*\frac{17}{\sqrt{47} }$ and $262+2.33*\frac{17}{\sqrt{47} }$

= (255.2223, 266.7777)