New York City is the most expensive city in the United States for lodging. The m
ID: 3350282 • Letter: N
Question
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55. a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)? b. What is the probability that a hotel room costs less than $140 per night (to 4 decimals)? c. What is the probability that a hotel room costs between $200 and $300 per night (to 4 decimals)? d. What is the cost of the 20% most expensive hotel rooms in New York City? Round up to the next dollar. $ or
Explanation / Answer
A)
Since =204 and =55
we have:
P ( X>225 )=P ( X>225204 )=P ( (X)/>(225204)/55)
Since Z=(x)/ and (225204)/55=0.38
we have:
P ( X>225 )=P ( Z>0.38 )
Use the standard normal table to conclude that:
P (Z>0.38)=0.352
B)
Since =204 and =55
we have:
P ( X<140 )=P ( X<140204 )=P ((X)/<(140204)/55)
Since( x)/=Z and (140204)/55=1.16
we have:
P (X<140)=P (Z<1.16)
P (Z<1.16)=0.123
C)
Since =204 and =55 we have:
P ( 200<X<300 )=P ( 200204< X<300204 )
=P ( (200204)/55<(X)/<((300204)/55)
Since Z=(x)/ , (200204)/55=0.07 and (300204)/55=1.75
we have:
P ( 200<X<300 )=P ( 0.07<Z<1.75 )
P ( 0.07<Z<1.75 )=0.4878
D) here mean 204, standard deviation=55
X= first top of the 20% costs $y say
P(x<=y)=0.80
So p(z<=(y-204)/55)
Here y=275.57 so 20% most expansive rooms cost 276 dollors or more