Question 1197939
Binomial Distribution:
Decade ago:
n = 202, p = .89 , 202*.89  and 202*.11 are  > 5
mean = 202*.89 = 179.89
sd = {{{sqrt(202*.89*.11)}}}= 4.447
NOW:
n = 202 ,  p = .16, 202*.16  and 202*.84  are > 5
Using the normal approximation and continuity correction factor.
(the continuity correction factor used as a Binomial Distribution is not continuous)
 mean =202*.16= 32.32
   sd = {{{sqrt(202*.16*.84)}}} = 5.21
Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
P(x  ≥  50 ) = normpdf( 49.5, 9999, 179.89,4.447)= 1
P(x  ≥  50 ) = normpdf(49.5, 9999,  32.32,5.21)= .0005