Question

A school sells 100 parking permits for a student parking lot with 80 spots and the students are independently/identically distributed. The students want to park in the lot 90% of the time. What is the probability that at least one student will not be able to park in the parking lot? Use the continuity method and central limit theorem to compute the probability.

Answer 1

n=

100

p=

0.9000

here mean of distribution=μ=np=		90.00
and standard deviation σ=sqrt(np(1-p))=		3.00
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

probability that at least one student will not be able to park in the parking lot =P(81 or more tried to park):

probability =P(X>80.5)=P(Z>(80.5-90)/3)=P(Z>-3.17)=1-P(Z<-3.17)=1-0.0008=0.9992