The margin of error for a sample proportion is given by

where:

is the z score associated with the confidence level, p is the sample prortion and n is the sample size.
We assume a confidence level of 95%, then
p = 52% = 0.52 and n = 2251
Therefore, margin of error =

The the interval that is likely to contain the true population proportion is between 49.9% and 54.1%.
Answer:
She made 792 dollars that week. I'm assuming it is just asking for the yearly commission she made. That would be 41,184 dollars a year.
Step-by-step explanation:
7200 x 0.11 = 792 Weekly
52 Weeks in a year
792 x 52 = 41184 Yearly (if that's what it's asking for)
60 /1 = 120 /2 = 180 /3
hope it helps