Question

the first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. if a sample of 15 random matches of the player is taken, the mean first-serve percentage is found to be 26.4%. what is the margin of error of the sample mean? 0.086% 0.533% 1.11% 2.22%

Answer 1

It is now possible to find the margin of error, the reason being that the confidence level is not given. However the 'standard error' of the sample mean can be found as follows:
$Standard\ error=\frac{\sigma}{ \sqrt{n} }=\frac{0.043}{ \sqrt{15} }=0.011103$
Converted to a percentage, we get 1.11% which is the third choice of answer.