Answer: The percentage of hotels in the city have a nightly cost of more than $200 is 21%
Step-by-step explanation:
Since the nightly cost of hotels in a certain city is normally distributed,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = the nightly cost of hotels.
µ = mean cost
σ = standard deviation
From the information given,
µ = $180.45
σ = $24.02
The probability that a hotel in the city has a nightly cost of more than $200 is expressed as
P(x > 200) = 1 - P(x ≤ 200)
For x = 200,
z = (200 - 180.45)/24.02 = 0.81
Looking at the normal distribution table, the probability corresponding to the z score is 0.79
Therefore,
P(x > 200) = 1 - 0.79 = 0.21
The percentage of hotels in the city have a nightly cost of more than $200 is
0.21 × 100 = 21%
