Let the lengths of pregnancies be X
X follows normal distribution with mean 268 and standard deviation 15 days
z=(X-269)/15
a. P(X>308)
z=(308-269)/15=2.6
thus:
P(X>308)=P(z>2.6)
=1-0.995
=0.005
b] Given that if the length of pregnancy is in lowest is 44%, then the baby is premature. We need to find the length that separates the premature babies from those who are not premature.
P(X<x)=0.44
P(Z<z)=0.44
z=-0.15
thus the value of x will be found as follows:
-0.05=(x-269)/15
-0.05(15)=x-269
-0.75=x-269
x=-0.75+269
x=268.78
The length that separates premature babies from those who are not premature is 268.78 days
The mean should be about the same, because the sample is random
<h2>
Explanation:</h2>
The rational expression is given by:

The denominator can't be zero, therefore:

Finally,<em> the values of x that make the rational expression undefined are </em>
<em></em>
<em>x = 7 and x = -7</em>
20 miles! :)
Good luck! Please make me brainliest if I was correct!