Question

63% of all Americans live in cities with population greater than 100,000 people. If 45 Americans are randomly selected, find the probability that a. Exactly 27 of them live in cities with population greater than 100,000 people. b. At most 27 of them live in cities with population greater than 100,000 people. c. At least 30 of them live in cities with population greater than 100,000 people. d. Between 23 and 29 (including 23 and 29) of them live in cities with population greater than 100,000 people.

Answer 1

n=

45

p=

0.6300

here mean of distribution=μ=np=		28.35
and standard deviation σ=sqrt(np(1-p))=		3.24
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

a)

probability =P(26.5<X<27.5)=P((26.5-28.35)/3.239)<Z<(27.5-28.35)/3.239)=P(-0.57<Z<-0.26)=0.3974-0.2843=0.1131

b)

probability =P(X<27.5)=(Z<(27.5-28.35)/3.239)=P(Z<-0.26)=0.3974

c)

probability =P(X>29.5)=P(Z>(29.5-28.35)/3.239)=P(Z>0.36)=1-P(Z<0.36)=1-0.6406=0.3594

d)

probability =P(22.5<X<29.5)=P((22.5-28.35)/3.239)<Z<(29.5-28.35)/3.239)=P(-1.81<Z<0.36)=0.6406-0.0351=0.6055