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

[jaffar]

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

 

[math_celebrity]

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