Question 1189368: Fiona walks from her house to a bus stop where she gets a bus to school. Her time, W minutes to walk to the us stop is normally distributed with W~N(12,3^2).
Fiona always leaves her house at 7:15. The first bus that she can get departs at 7:30.
a) Find the probability that it will take Fiona between 15 to 30 minutes to walk to the bus stop.
The length of time, B minutes of the bus journey to Fiona's school is normally distributed with B~N (50,σ^2). The probability that the bus journey takes less than 60 minutes is 0.941.
Find σ and the probability that the bus journey takes less than 45 minutes,
If Fiona misses the first us there is a second bus which departs at 7:45. She must arrive at school y 8:30. Fiona will not arrive if she misses these buses. The variables W and B are independent.
Find the probability that she will be on time.
This year if she goes to school on 183 days, find the numbers of day she's expected to be on time.
I'm really confused on how to solve this, and where to start. Please explain step-by-step if possible.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! mean is 12 sd=3, the sqrt of the variance of 9.
z=(x-mean)/sd
a. z is between 1 and 6 sd with probability of 0.1587. This is the normalized time of 15 minutes (+1 sd and 30 minutes +6 sd).
b. Z < 60 is probability of 0.941, or z=1.563, taking invnorm of 0.941 VARS2 then 3 invnorm and then (invnorm 0.941,0,1) ENTER.
1.563=(60-50)/sd
sd=10/1.563=6.397 minutes
The bus has mean 50 minutes and sd 6.397 min. So under 60 minutes is under 1.56 sd and that is where the probability of 0.941 comes in.
probability < 45 minutes is z <-5/6.397=-0.7816 and that probability less than that is 0.2172
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probability she will be on time is: makes the first bus with 0.8413 probability (making it under 15 minutes or under 1 sd). Then the probability the bus gets there is in 1 hour is 0.941. The probability she gets there on time AND the bus is on time is the product of the probabilities or 0.7917.
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the other issue is that it takes her more than 15 minutes but fewer than 30 minutes (between 1 and 6 sd). that probability is 0.1587 * the probability the bus gets there under 45 minutes (0.2172). That joint probability is the product of both of those, because they are independent, and is 0.0345.
The sum of both of those probabilities is 0.8262
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Even if she makes the first bus, she has a 5.9% chance right there of being late. If she misses the second bus (>2 sd or 0.0228 probability), she won't make it at all. So right there her probability is 8.1% chance of being late.
She has only an 0.8413 probability of making the first bus, and there is not much likelihood of making it on time with the second bus (0.2172). To make school arriving late and having the bus get there faster than usual is the product of two small probabilities.
-multiply the probability of 0.8262*183=151.19 or 151 days on time.
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