Question 670700: 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 = 17.2 minutes with σ = 4.4 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? Hint: Compute P(x > 25|x > 18).
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean = 17.5
standard deviation is 4.4
you need to find the z-scores for 15 minutes, 18 minutes, 20 minutes and 25 minutes.
the z-score is equal to (x-m)/s
x is the raw score.
m is the mean.
s is the standard deviation.
exampl:
for 15 minutes, the z-score is (15 - 17.5) / 4.4
the z-scores are:
15 minutes = -.5
18 minutes = .18182
20 minutes = .63636
25 minutes = 1.77273
all rounded to the nearest 5 decimal places.
you then look up in the normal distribution tables to find the probability of getting a z-score greater than the indicated z-score.
from the tables, you get:
p(z > -.5) = .69146
p(z > .42786) = .42786
p(z > .26227) = .26227
p(z > 25) = .03814
translated to raw scores, these probabilities are the same as:
p(x > 15) = .69146
p(x > 18) = .42786
p(x > 20) = .26227
p(x > 25) = .03814
now you need to find:
p(x > 20 | x > 15)
p(x > 25 | x > 18)
the general formula is p(a | b) = p(a intersect b) / p(b)
translated to the first problem, this formula becomes:
p(x > 20 | x > 15) = p(x > 20 intersect x > 15) / p(x > 15).
p(x > 20 intersect x > 15) means that you have to wait greater than 15 minutes and you have to wait greater than 20 minutes at the same time. That happens when you are waiting greater than 20 minutes and not before.
Therefore, p(x > 20 intersect x > 15) must be p(x > 20).
that's my take on the intersect part of the equation.
the equation therefore becomes:
p(x > 20 | x > 15) = p(x > 20 intersect . > 15) / p(x > 15) = .26277 / .69146 which becomes equal to .37930
similarly, for the second problem, the formula becomes:
p(x > 25 | x > 18) = p(x > 25 intersect x > 18) / p(x > 18).
since p(x > 25 intersect x > 18) can only occur when x > 25, then this is equal to p(x > 25) which is equal to .03814.
the formula therefore becomes:
p(x > 25 | x > 18) = .03814 / .42786 = .08914
the formulas are given.
the main problem is in the interpretation of what constitutes a intersect b.
the logic looks sound so i'd go with this answer unless you have an answer from somebody else that you like better.
if it was me, i'd go with this one.
the alternate way to look at it is this.
suppose you went to the restaurant 100,000 times.
you would have had to wait greater than 15 minutes for 69,146 of those times.
you would have had to wait greater than 20 minutes for 26,227 of those times.
given that you had to wait greater than 15 minutes means your universe from which you select from is now 69,146 rather than 100,000.
the probability of waiting greater than 20 minutes given that you had to wait greater than 15 minutes becomes 26,227 / 69,146 which is equal to .37930
this means that, of the 69,146 times you had to wait greater than 15 minutes, you had to wait greater than 20 minutes 37.930 % of those times.
|
|
|
