Question 1197940
Binomial Distribution:
n = 136,  p = .38
 136*.38  and 136*.62  are > 5
Using the normal approximation and continuity correction factor.
(the continuity correction factor used as a Binomial Distribution is not continuous)
 mean =136*.38= 51.68
   sd = {{{sqrt(136*.38*.62)}}} = 5.66
Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
P(x  ≥  68 ) = normpdf(67.5,9999,  51.68, 5.66)= .0026
P(x  <  45 ) = normpdf(-9999, 44.5,  51.68, 5.66)= .1023
P( 40 ≤ x ≤ 64) = normcdf( 39.5,64.5,  51.68, 5.66) = .9725
P(x > 80) = normpdf(80.5,9999, 51.68, 5.66) = .0000002