Question

a local hotel has 600 available rooms. At current prices, it fircats that the mean demand for hotel rooms for the thanksgiving weekend will be 570 rooms with a standard deviation of 30. let X be the drmand for hotel rooms and assume that it has a normal distribution as suggested fron the analysis of historical data.

what is the probability that at least 90% of the rooms get rented? That is, what is the probability that the demand for rooms is 540 or more?

answr in decimals pls

What is the 95th percentile of the distribution? answer in terms of numbers of rooms

Answer 1

Given, mean = µ = 570 and standard deviation = σ = 30

Let the number of rooms rented be x.

Probability that 90% of the rooms are rented:

P ( x ≥ 540 ) = P [(x-µ)/σ ≥ (540-570)/30] = P ( z ≥ -1 ) = P( -1 ≤ z < ∞ ) = P ( -1 ≤ z ≤ 0 ) + P ( 0 ≤ z < ∞ )

= (0.5-0.1587) + 0.5 = 0.8413 [values from z-table]

So, the probability that 90% of the rooms are rented is 0.8413

The percentile for a normal ditribution is given by:

x = µ + zσ

Here µ = 570, σ = 30 and for 95th percentile z = 1.645, so

x = 570 + (1.645*30) = 570 + 49.35 = 619.35 ~ 619

So, the 95th percentile of this distribution is 619 rooms