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
Which question and if you have the name of the assignment or it has a copy right at the bottom you can most likely find it online hope that helps
We need the graphing tool.
Answer:
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