Question

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night.† Assume that room rates are normally distributed with a standard deviation of $55.

(a)

What is the probability that a hotel room costs $265 or more per night? (Round your answer to four decimal places.)

(b)

What is the probability that a hotel room costs less than $110 per night? (Round your answer to four decimal places.)

(c)

What is the probability that a hotel room costs between $210 and $290 per night? (Round your answer to four decimal places.)

(d)

What is the cost in dollars of the 20% most expensive hotel rooms in New York City? (Round your answer to the nearest cent.)

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	204
std deviation   =σ=	55.0000

a)

probability that a hotel room costs $265 or more per night :

probability =

P(X>265)

=

P(Z>1.11)=

1-P(Z<1.11)=

1-0.8665=

0.1335

b)

probability that a hotel room costs less than $110 per night :

probability =

P(X<110)

=

P(Z<-1.71)=

0.0436

c)

probability that a hotel room costs between $210 and $290 per night :

probability =

P(210<X<290)

=

P(0.11<Z<1.56)=

0.9406-0.5438=

0.3968

d)

for 80th percentile critical value of z=		0.84
therefore corresponding value=mean+z*std deviation=			250.29