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 .

(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 and

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

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

So 98% of pregnancies would lie between

and

= (255.2223, 266.7777)

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

the correct answer is b. negative exponential growth.

step-by-step explanation:

you can tell it's an exponential function because it decreases rapidly then levels out and never touches the x-axis.

it is negative because the y-axis decreases as the x-axis increases.

therefore it is a negative exponential growth graph.

hope i !