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 $255 or more per night? (Round your answer to four decimal places.)

(b)

What is the probability that a hotel room costs less than $130 per night? (Round your answer to four decimal places.)

(c)

What is the probability that a hotel room costs between $200 and $280 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

Solution :

Given that ,

mean = = 204

standard deviation = = 55

a)

P(x 255) = 1 - P(x   255)

= 1 - P((x - ) / (255 - 204) / 55)

= 1 -  P(z 0.93)  

= 1 - 0.8238   

= 0.1762

Probability = 0.1762

b)

P(x < 130) = P((x - ) / < (130 - 204) / 55)

= P(z <-1.35 )

= 0.0885

Probability = 0.0885

c)

P(200 < x < 280) = P((200 - 204)/ 55) < (x - ) /  < (280 - 204) / 55) )

= P(-0.07 < z < 1.38)

= P(z < 1.38) - P(z < -0.07)

= 0.9162 - 0.4721

= 0.4441

Probability = 0.4441

d)

P(Z > z ) = 20%

1 - P(Z < z) = 0.20

P(Z < z ) = 0.80

P(Z < 0.84) = 0.80

Using z-score formula,

X = z* +

= 0.84 * 55 + 204

= 250.20