Question

An individual house in a community spends an average of $400 per month on home maintenance with a standard deviation of $45.

a) A community has 70 houses. What is the probability that the community pays more than $27750 on home maintenance?

b) Among 80 communities with 70 houses each, what is the probability that at least 30 of them pay less than $27750?

Answer 1

a)

expected value of 70 house =70*400=28000

standard deviation of total =45*sqrt(70)=376.497

probability that the community pays more than $27750 on home maintenance:

probability =P(X>27750)=P(Z>(27750-28000)/376.497)=P(Z>-0.66)=1-P(Z<-0.66)=1-0.2546=0.7454

b)

n=

80

p=

0.7454

here mean of distribution=μ=np=		59.63
and standard deviation σ=sqrt(np(1-p))=		3.90
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

probability that at least 30 of them pay less than $27750 =P(at most 50 pays more than 27750):

probability =P(X<50.5)=(Z<(50.5-59.632)/3.896)=P(Z<-2.34)=0.0096