Statistics: discrete probability and binomial distribution

Question

70% of all students at a college still need to take another math class. If 45 students are randomly selected, find the probability that :

a) At least 33 of them need to take another math class

b) Between 30 and 34 (including 30 and 34) of them need to take another math class

Answer 1

np=45x.7= 31.5=mean

npq= 31.5x.3= 9.45

sqrnpq=sd= sqr9.45=about 3.0741= standard deviation

z=33-31.5)/3.0741=about 1.5/3 = 1/2

use z tables

Pr(x>33)= .309

about 30% chance at least 33 need to take another math class

30 to 34

z = (34-31.5)/3 = about 2.5/3 = 5/6 = about .667

z = (30-31.5)/3 = about -1.5/3 = -1/2 = -.500

Pr(30 to 34) = about .439

= about 44% between 33 and 34 need another math class

using the binomial to approximate the normal distribution

n = number=45

p = .7 = 70%

q = 1-p= .3 = 30%

np= mean

sqr(npq) = standard deviation

npq = variance