Question

According to a​ study, the median time a patient waits to see a doctor in an emergency room is 30 minutes. Consider an emergency room on a day when 150 patients visit.

a.What is the probability that more than half will wait more than 30​ minutes?

b.What is the probability that more than 80 will wait more than 30​ minutes?

c.What is the probability that more than 60 but less than 90 will wait more than 30​ minutes?

Answer 1

probability a patient waits more than 30 minutes:

P(X>30)=1-P(X<30)=1-(1-exp(-30/30))=0.3679

here for binomial distribution parameter n=150 and p=0.3679

mean of distribution=μ=np=		55.1850
and standard deviation σ=sqrt(np(1-p))=		5.9061
for normal distribution z score =(X-μ)/σx
since np and n(1-p) both are greater than 5, we can use normal approximation of binomial distribution
therefore from normal approximation of binomial distribution and continuity correction:

a)

probability that more than half will wait more than 30​ minutes :

probability =P(X>75.5)=P(Z>(75.5-55.185)/5.906)=P(Z>3.44)=1-P(Z<3.44)=1-0.9997=0.0003

b)

probability that more than 80 will wait more than 30​ minutes :

probability =P(X>80.5)=P(Z>(80.5-55.185)/5.906)=P(Z>4.29)=1-P(Z<4.29)=1-1.0000=0.0000

c) probability that more than 60 but less than 90 will wait more than 30​ minutes :

probability =P(60.5<X<89.5)=P((60.5-55.185)/5.906)<Z<(89.5-55.185)/5.906)=P(0.9<Z<5.81)=1-0.8159=0.1841