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

Discussion in 'Calculator Requests' started by jaffar, Nov 14, 2016.

1. ### jaffarNew Member

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
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