The length of the human pregnancy is not fixed. It is known that it varies according to a distribution which is roughly normal, with a mean of 266 days, and a standard deviation of 16 days. a. Fill in the curve below with the % and X-axis b. Approximately what percent of pregnancy are between 234 and 298 days c. Approximately what percent of pregnancy are between 250 and 314 days d. Approximately what percent of pregnancy are below 218

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part b

Let X the random variable that represent the length of human pregnancy of a population, and for this case we know that:

Where [tex]\mu=266[/tex] and [tex]\sigma=16[/tex]

We are interested on this probability

[tex]P(234<X<298)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(234<X<298)=P(\frac{234-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{298-\mu}{\sigma})=P(\frac{234-266}{16}<Z<\frac{298-266}{16})=P(-2<z<2)[/tex]

And we can find this probability with this difference:

[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]

An we know using the graph in part a that this area correspond to 0.95 or 95%

Part c

[tex]P(250<X<314)=P(\frac{250-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{314-\mu}{\sigma})=P(\frac{250-266}{16}<Z<\frac{314-266}{16})=P(-1<z<3)[/tex]

And we can find this probability with this difference:

[tex]P(-1<z<3)=P(z<3)-P(z<-1)[/tex]

An we know using the graph in part a that this area correspond to 0.95 or 34+34+13.5+2.35%=83.85%

Part d

We want this probability:

[tex] P(X<218)[/tex]

And using the z score we got:

[tex]P(X<218) = P(Z< \frac{218-266}{16}) = P(Z<-3) [/tex]

And that correspond to approximately 0.15%