Answer:
68.268%.
Explanation:
Let \mu be the mean pregnancy length.
The given standard deviation,
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

z-score for this range is:


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
to z=1 is 0.84134
and the area from
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%.