standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and margin of error 128

Standard Error (margin of Error) = Standard Deviation / sqrt(n) 128 = 545/sqrt(n) Cross multiply: 128sqrt(n) = 545 Divide by 128 sqrt(n) = 4.2578125 Square both sides: n = 18.1289672852 But we need an integer, so the answer is 19