Question

In New York City the average hotel room costs $204 per night, with a standard deviation of $55. Suppose that you walk into a random hotel and learn their rate is $155 per night. If the rates are normally distributed, what is the probability that if you leave and walk into another random hotel, they will have a lower rate

Answer 1

The value is
`P( X <  155)   = 0.18649`

Explanation:

From the question we are told that

The mean is
`\mu = \$ 204`

The standard deviation is
`\sigma = \$ 55`

Generally the probability that another hotel will a rate lower than $155 per night is mathematically represented as

`P( X < 155) = P( (X - \mu )/(\sigma) < (155 - 204 )/( 55) )`

`(X -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \ X )`

=>
`P( X < 155) = P( Z < -0.8909 )`

From the z-table the area under the normal curve corresponding to -0.8909, towards the left is

`P( Z <  -0.8909 ) =0.18649`

=>
`P( X <  155)   = 0.18649`