Respuesta :
Answer:
68.268%.
Step-by-step explanation:
Let \mu be the mean pregnancy length.
The given standard deviation, [tex]\sigma=7[/tex] days
The z-score:
z=\frac{x-\mu}{\sigma}
where x is any pregnancy length.
The range of x within 7 days of the mean pregnancy length is
[tex]\mu-7<x<\mu+7.[/tex]
z-score for this range is:
[tex]\frac{\mu-7-\mu}{\sigma}<z<\frac{\mu+7-\mu}{\sigma}[/tex]
[tex]\Rightarrow -1<z<1[/tex]
So, the area on the normally distributed curve for the above range of z-score will give the expected percentage of birds to occur within 7 days of the mean pregnancy length.
Now, from the z-score table:
The area from [tex]z = -\infty[/tex] to z=1 is 0.84134
and the area from [tex]z = -\infty[/tex] to z=-1 is 0.15866.
So, the area for -1<z<1
=0.84134 - 0.15866
=0.68268
Hence, the percentage of births would be expected to occur within 7 days of the mean pregnancy length=68.268%.
