New York City is the most expensive city in the United States for lodging. The room rate is $204 per night (USA Today, April 30, 2012). Assu

Question

New York City is the most expensive city in the United States for lodging. The room rate is $204 per night (USA Today, April 30, 2012). Assume that room ratesa mally distributed with a standard deviation of $55. a. What is the probability that a hotel room costs $225 or more per ni b. What is the probability that a hotel room costs less than $140 per n c. What is the probability that a hotel room costs between $200 and $300 per ni me 25. are nor ght? ight? d. What is the cost of the most expensive 20% of hotel rooms in New York City?

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Rose 3 weeks 2021-10-01T18:33:56+00:00 1 Answer 0

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    2021-10-01T18:35:04+00:00

    Answer:

    a. P=0.35

    b. P=0.12

    c. P=0.49

    d. Cost: more than $249 per night.

    Step-by-step explanation:

    We assume a normal distribution with mean $204 and s.d of $55.

    To calculate the probability we calculate the z-value for each case

    a. What is the probability that a hotel room costs $225 or more

    z=\frac{X-\mu}{\sigma}=\frac{225-204}{55}= \frac{21}{55}= 0.382\\\\P(X>225)=P(z>0.382)=0.35

    b. What is the probability that a hotel room costs less than $140

    z=\frac{140-204}{55}= \frac{-64}{55}=-1.164\\\\P(X<140)=P(z<-1.164)=0.12

    c. What is the probability that a hotel room costs between $200 and $300

    z=\frac{200-204}{55}= \frac{-4}{55}=-0.073\\\\P(X<200)=P(z<-0.073)=0.47\\\\\\z=\frac{300-204}{55}= \frac{96}{55}=1.745\\\\P(X<300)=P(z<1.75)=0.96\\\\\\P(200<X<300)=P(X<300)-P(X<200)=0.96-0.47=0.49

    d. What is the cost of the most expensive 20% of hotel rooms in New York City?

    In this case, we have to estimate z’ so thats P(z>z’)=0.2. This value is z=0.841.

    Then, the value of X should be:

    X=\mu+z*\sigma=204+0.815*55=204+45=249

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45:7+7-4:2-5:5*4+35:2 =? ( )