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

Answer 1

A)

Since μ=204 and σ=55

we have:

P ( X>225 )=P ( X−μ>225−204 )=P ( (X−μ)/σ>(225−204)/55)

Since Z=(x−μ)/σ and (225−204)/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 μ=204 and σ=55

we have:

P ( X<140 )=P ( X−μ<140−204 )=P ((X−μ)/σ<(140−204)/55)

Since( x−μ)/σ=Z and (140−204)/55=−1.16

we have:

P (X<140)=P (Z<−1.16)

P (Z<−1.16)=0.123

C)

Since μ=204 and σ=55 we have:

P ( 200<X<300 )=P ( 200−204< X−μ<300−204 )

=P ( (200−204)/55<(X−μ)/σ<((300−204)/55)

Since Z=(x−μ)/σ , (200−204)/55=−0.07 and (300−204)/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