Question

Suppose you want to eat lunch at a popular restaurant. The restaurant does not take reservations, so there is usually a waiting time before you can be seated. Let x represent the length of time waiting to be seated. From past experience, you know that the mean waiting time is

μ = 17.8 minutes with σ = 4.3 minutes. You assume that the x distribution is approximately normal. (Round your answers to four decimal places.)
(a) What is the probability that the waiting time will exceed 20 minutes, given that it has exceeded 15 minutes? Hint: Compute P(x > 20|x > 15).
(b) What is the probability that the waiting time will exceed 25 minutes, given that it has exceeded 18 minutes?

Answer 1

a)
`P(X>20 |X>15) =(P(X>20))/(P(X>15)) =(0.305)/(0.742)=0.411`

b)
`P(X>25 |X>18) =(P(X>25))/(P(X>18)) =(0.048)/(0.481)=0.0997`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the interpupillary distance (the distance between the pupils of the left and right eyes) of a population, and for this case we know the distribution for X is given by:

`X \sim N(17.8,4.3)`

Where
`\mu=17.8` and
`\sigma=4.3`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

On this case
`\bar X \sim N(72,(6)/(√(27)))`

Part a

(a) What is the probability that the waiting time will exceed 20 minutes, given that it has exceeded 15 minutes?

`P(X>20 |X>15) =(P(X>20 and X>15))/(P(X>15))=(P(X>20))/(P(X>15))=(1-P(X<20))/(1-P(X<15))`

We can find the inidivual probabilities like this:

`P(X>15)=1-P(X<15) = 1-P(Z<(15-17.8)/(4.3))=1-P(Z<-0.65)=1-0.258=0.742`

`P(X>20)=1-P(X<20) = 1-P(Z<(20-17.8)/(4.3))=1-P(Z<0.511)=1-0.695=0.305`

And then we can replace

`P(X>20 |X>15) =(P(X>20))/(P(X>15)) =(0.305)/(0.742)=0.411`

Part b

(b) What is the probability that the waiting time will exceed 25 minutes, given that it has exceeded 18 minutes?

`P(X>25 |X>18) =(P(X>25 and X>18))/(P(X>18))=(P(X>25))/(P(X>18))=(1-P(X<25))/(1-P(X<18))`

We can find the inidivual probabilities like this:

`P(X>18)=1-P(X<18) = 1-P(Z<(18-17.8)/(4.3))=1-P(Z<0.0465)=1-0.519=0.481`

`P(X>25)=1-P(X<25) = 1-P(Z<(25-17.8)/(4.3))=1-P(Z<1.67)=1-0.952=0.048`

And then we can replace

`P(X>25 |X>18) =(P(X>25))/(P(X>18)) =(0.048)/(0.481)=0.0997`