Answer:
A. P(x>300)=0.0166
Step-by-step explanation:
A. The problem tell us we are working in a Normal Distribution with mean 268 and STD of 15, and we are asked for the probability that a pregnancy last at least 300 days. The probability is not meant to be found directly (it´s too difficult to find it in "that" distribution), there´s a process called "Normalize" which is used to transfer the problem from an specific Normal Distribution to a Standard Normal Distribution (with mean 0 and STD 1) where we are able to find probabilities easier as follows:
P(x<a)= P( (x-μ)/σ < (a-μ)/σ ) = P(z<b)
Where x,a,μ,σ are data in the specific Normal Distribution (a the value that we want its probability, μ the mean and σ the STD) and z,b are equivalent data to x,a but they are from a Standard Normal Distribution.
It´s useful to do this because in a Standard Normal Distribution we are able to find probabilities easier. A way to find those values, is to use a chart (it´s attached to this answer) with the probabilities that works like this:
P(z<b)= (u,v)
Where (u,v) are the coordinates of the value of the probability with u=whole part and first decimal of b, and v=second decimal of b. Keep in mind that it only works with probabilities UNDER a value ("<")
We will use this problem as an example.
First we write the probability we want and "Normalize" it:
P(x>300) = P( (x-μ)/σ > (300-268)/15 ) = P(z>2,133...)
But the chart only works with "<", so we rewrite it with a property of probability:
P(z>2,133...) = 1 - P(z<2,133...)
And now, we are able to look for the value of this probability:
P(z<2,1333...) = the place with coordinates (2,1 , 4 (because it´s periodic))
P(z<2,1333...) =0.9834
P(z>2,133...) = 1 - 0.9834 = 0.0166
But we know P(x>300) = P(z>2,133...), so our answer will be:
P(x>300) = 0.0166
