Answer:

And we want this probability:

And using the normal standard distribution or excel we got:

And the best option would be:
D 0.0233
Explanation:
For this case we have the following parameters:

And we select a sample size of n=13. And we want to find this probability:

We can assume that the distribution for this case is normal and then we can use the z score formula given by:

And if we replace the data given we got:

And we want this probability:

And using the normal standard distribution or excel we got:

And the best option would be:
D 0.0233