Respuesta :
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
[tex]z=\frac{X-\mu}{\sigma}=\frac{225-204}{55}= \frac{21}{55}= 0.382\\\\P(X>225)=P(z>0.382)=0.35[/tex]
b. What is the probability that a hotel room costs less than $140
[tex]z=\frac{140-204}{55}= \frac{-64}{55}=-1.164\\\\P(X<140)=P(z<-1.164)=0.12[/tex]
c. What is the probability that a hotel room costs between $200 and $300
[tex]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[/tex]
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:
[tex]X=\mu+z*\sigma=204+0.815*55=204+45=249[/tex]
