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?

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,

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:

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

isi

step-by-step explanation:

use that smart brain

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.