P(a)=.60, p(b)=.20, and p(a and b)=.15

Assemble the proof by dragging tiles to the statements and reasons column

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? a. 0.086% b. 0.533% c. 1.11% d. 2.22%

|p| > 3 {-3, 3} {p|-3 < p < 3} {p|p < -3 or p > 3}