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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B225-204%7D%7B55%7D%3D%20%5Cfrac%7B21%7D%7B55%7D%3D%200.382%5C%5C%5C%5CP%28X%3E225%29%3DP%28z%3E0.382%29%3D0.35)
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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B140-204%7D%7B55%7D%3D%20%5Cfrac%7B-64%7D%7B55%7D%3D-1.164%5C%5C%5C%5CP%28X%3C140%29%3DP%28z%3C-1.164%29%3D0.12)
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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B200-204%7D%7B55%7D%3D%20%5Cfrac%7B-4%7D%7B55%7D%3D-0.073%5C%5C%5C%5CP%28X%3C200%29%3DP%28z%3C-0.073%29%3D0.47%5C%5C%5C%5C%5C%5Cz%3D%5Cfrac%7B300-204%7D%7B55%7D%3D%20%5Cfrac%7B96%7D%7B55%7D%3D1.745%5C%5C%5C%5CP%28X%3C300%29%3DP%28z%3C1.75%29%3D0.96%5C%5C%5C%5C%5C%5CP%28200%3CX%3C300%29%3DP%28X%3C300%29-P%28X%3C200%29%3D0.96-0.47%3D0.49)
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](https://tex.z-dn.net/?f=X%3D%5Cmu%2Bz%2A%5Csigma%3D204%2B0.815%2A55%3D204%2B45%3D249)