Question 1178691
What we want is 

{{{abs(p[1] - p) <= z[0.96]*sqrt((p(1-p))/n) = 0.05 }}}, and 

==> {{{ 1.751*sqrt((0.27*0.63)/n) = 0.05 }}}

==> {{{sqrt((0.27*0.63)/n) = 0.05/1.751 }}}

==> {{{(0.27*0.63)/n = (0.05/1.751)^2 }}}

==> {{{n/(0.27*0.63) = (1.751/0.05)^2}}}


==> {{{n = 0.27*0.63*(1.751/0.05)^2 = 208.61}}}, to 2 d.p.

This then implies to choose a sample with a size of {{{highlight(209)}}}.

(Normal probabilities were taken from https://stattrek.com/online-calculator/normal.aspx.)