Fred-to-bus and bus-to-bus average times

a)fred visits blotchville again. he finds that the city has installed an electronic display at the bus stop, showing the time when the previous bus arrived. the times between arrivals of buses are still independent exponentials with mean 10 minutes. fred waits for the next bus, and then records the time between that bus and the previous bus. on average, what length of time between buses does he see?

(b) fred then visits blunderville, where the times between buses are also 10 minutes on average, and independent. yet to his dismay, he finds that on average he has to wait more than 1 hour for the next bus when he arrives at the bus stop! how is it possible that the average fred-to-bus time is greater than the average bus-to-bus time even though fred arrives at some time between two bus arrivals? explain this intuitively, and construct a specific discrete distribution for the times between buses showing that this is possible

your exact question is kind of difficult to understand, so maybe an example will explain what you're supposed to do? i'll work through the first problem on the worksheet.

suppose you have a tray of 652 brownies, and you have 5 friends that you want to feed without picking favorites. how many brownies can you dole out to each of your friends in equal amounts? how many will be left over for you to eat?

what you're doing here is finding a way to arrange the number 652 as the sum of integer multiple of 5 and some remainder:

here ("q" for quotient) is the number of brownies each of your friends get, and is the remainder. that's the basic idea behind division.

we break up the number 652 into parts so that the leading digit stands on its own:

652 = 600 + 52

first we check if it's possible to hand out exactly 100 brownies to each friend. with 5 friends, this means we want to know if we can take out 500 brownies from the tray. we can do this 1 time because we have 600 brownies to take from, and when we do that we're left with 100 + 52 brownies to divvy up. so we know that

652 = 1*500 + 152

next we want to know if it's possible to hand out exactly 10 brownies per friend. if we can, we have to remove 50 brownies from the tray each time until we can't. we have 152 brownies left, and we know that 3(50) = 150, so

152 = 3*50 + 2

now we have 2 brownies left in the tray. is it possible to hand 1 out to each of your 5 friends? no, you'd need a minimum of 5 brownies to do that. so the last 2 are yours.

putting everything together, we found that

652 = 1*500 + 3*50 + 2

and we can rearrange this to write

652 = 5*100 + 5*30 + 2

652 = 5(100 + 30) + 2

652 = 5*130 + 2

so and .

you might be more accustomed to the table depicting the algorithm. that's just a compact way of organizing what we've done above. you can consult the image i've attached below.

step-by-step explanation:

the line y=x is the line where the x and y coordinates of each point, are equal.

so this line is the line through the origin (0, 0), which makes an angle of 45° with the x-axis.

we draw triangle abc. to reflect this triangle with respect to the line y=x, we draw perpendiculars from the points a, b and c to to line y=x, and extend the perpendiculars so that the lint y=x is a perpendicular bisector of these line segments. 

we see that a(5, 3) is mapped to a'(3, 5)

and all the other points p(a, b) of the original triangle are mapped to p'(b,a)

the last statement is true for each point x and its reflection x', also it is true that the line y=x is the perpendicular bisector of xx' and yy'.

to prove these, check the second picture.

let the coordinates of x be (a, b), we want to find the coordinates of its reflection point with respect to the line y=x, point x'.

let m be the point on the line y=x, such that xm=mx'

let n be the point (b, b) which surely is on the line.

now, x'm=mx, nm is common and m(x'mn)=m(xmn)=90°, so by side-angle-side congruence postulate, the triangles x'mn and xmn

  are congruent, so nx=nx'

clearly the x coordinate of x' is b, so the y coordinate of x' is 

b+nx'=b+nx=a

so the coordinates of x' are (b, a)

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