Respuesta :
Answer:
(a) The probability that all the hotel's rooms will be rented is 0.1587.
(b) The probability that 50 or more rooms will not be rented is 0.2514.
Step-by-step explanation:
We are given that the Broadmoor Hotel in Colorado Springs contains 700 rooms and is on the 2004 Gold List.
Suppose Broadmoor's marketing group forecasts a mean demand of 670 rooms for the coming weekend. Assume that demand for the upcoming weekend is normally distributed with a standard deviation of 30.
Let X = demand for rooms in the hotel
So, X ~ Normal([tex]\mu=670,\sigma^{2} =30^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean demand for the rooms = 670
[tex]\sigma[/tex] = standard deviation = 30
(a) The probability that all the hotel's rooms will be rented means that the demand is at least 700 = P(X [tex]\geq[/tex] 700)
P(X [tex]\geq[/tex] 700) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{700-670}{30}[/tex] ) = P(Z [tex]\geq[/tex] 1) = 1 - P(Z < 1)
= 1 - 0.8413 = 0.1587
The above probability is calculated by looking at the value of x = 1 in the z table which has an area of 0.8413.
(b) The probability that 50 or more rooms will not be rented is given by = P(X [tex]\leq[/tex] 650)
P(X [tex]\leq[/tex] 650) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{650-670}{30}[/tex] ) = P(Z [tex]\leq[/tex] -0.67) = 1 - P(Z < 0.67)
= 1 - 0.7486 = 0.2514
The above probability is calculated by looking at the value of x = 0.67 in the z table which has an area of 0.7486.
