The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. what is the probability that a pregnancy last at least 300 days? round to four decimal places.

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

y=mx+