At the beginning of winter, a homeowner estimates that the...

At the beginning of winter, a homeowner estimates that the probability is 0.4 that his total heating bill for the three winter months will be less than $380. He also estimated that the probability is 0.6 that the total bill will be less than $460. Assume that the total 3 month heating bills follow a uniform distribution. a.) What is the probability that the total bill will be between $380 and $460? b.) What is the probability that the total bill will be less than $400?

Transcript

00:01 We're looking at a uniform distribution here.

00:03 We don't know the interval, a and b.

00:06 So i'll start by drawing our distribution.

00:09 We have a and we have b, and every point between the two has an equal probability.

00:17 We don't know what they are.

00:18 We do know the probability of less than $380 is 0 .4, and that less than $460 is 0 .6.

00:27 So if i mark those on here, there'll be somewhere here.

00:30 0 .4 is 380, 0 .6 is 460.

00:38 So i know that this area here is 0 .4, i know this total area here is 0 .6.

00:45 What do we do now? well, what's the formula for getting probability out of a uniform distribution? it's getting the length of your interval divided by the total width of this rectangle.

00:58 It's like comparing two rectangles.

01:00 If we have this small rectangle here of size 0 .4, we're comparing it to the big rectangle of size 1, since it's a probability curve.

01:08 So if we compare the areas of these rectangles, both are going to be length times height, but they have the same height, so it's really just comparing length.

01:19 What's the length of the overall rectangle? well, it's b minus a.

01:23 What's the length of this smaller rectangle? well, it's 380 minus a.

01:29 And its area is 0 .4.

01:35 So we do know this.

01:38 We can do the same thing for 460.

01:41 460 minus a over b minus a is 0 .6.

01:45 Now i have two equations and i have two unknown variables.

01:49 I can work with this.

01:51 I can solve these simultaneously and i'll know a and b.

01:54 So that's what we're going to do next.

01:58 So my first trick is going to be rearranging these since i've got b minus a.

02:03 I want to make each of these have b minus a on the right hand side.

02:08 There are of course many ways you can solve this.

02:10 This is just going to be what i do.

02:12 So my first equation is going to be 380 minus a equals 0 .4, b minus a, so i'll divide by 0 .4, b minus a.

02:22 My second equation, 460 minus a over 0 .6 is b minus a.

02:30 Now i've got the right hand sides the same.

02:33 I can make these equal to each other.

02:35 So i've got 380, and i'll just divide by 0 .4 here.

02:41 And let's see what's 1 minus, 1 divided by 0 .4, that's 2 .5.

02:47 So 2 .5a is equal to 460 over 0 .6 minus and what's 1 divided by 0 .6.

02:59 Okay, 1 .6 recurring, which i might prefer to leave as 1 over 0 .6 a.

03:07 Right, so i'm going to mix these around a bit, get some more information.

03:12 Actually, i will leave this as a over 0 .6, to make my life a bit easier.

03:18 So i've got 380 divided by 0 .4, which is 950, and i'm going to take away this from both sides.

03:30 So i've got 460 divided by 0 .6, which is, of course, kind of horrible, 766 .6 recurring...

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