Solution :
Given that,
standard deviation = = 759.00
Margin of error = E = 19
At 80% confidence level the z is ,
Z/2 = Z0.1 = 1.282
sample size = n = [Z/2* / E] 2
n = ( 1.282* 759 / 19 )2
n =2622.722852
Sample size = n =2623
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