My notes for this chapter suck, so I

[billbennett770]

Homework Statement

Given the continuous probability density function f(x) = (x-2)/18 2 ≤ x ≤ 8, find P(6<x≤8)?

Homework Equations

For a continuous random variable, if you have a pdf f(x) on the interval a to b,

P(c≤x≤d)=∫dcf(x)dx
E(X)=μ=∫baxf(x)dx
Var(X)=∫ba(x−μ)2f(x)dx

For this problem, you are given with pdf f(x) and the interval is given. So you need to evaluate the integral.

The Attempt at a Solution

∫baf(x)dx= ∫8,2((x-2)/18))... I honestly lose it from here.

 

[Simon Bridge]

$f(x)=\frac{x-2}{18}: 2\leq x \leq 8$

You have to evaluate: $$p(6<x\leq 8)=\int_6^8 \frac{x-2}{18}\; dx$$ ... you had the wrong lower limit.
Show me your best attempt from there.

 

[SteamKing]

Homework Statement

Given the continuous probability density function f(x) = (x-2)/18 2 ≤ x ≤ 8, find P(6<x≤8)?

Homework Equations

For a continuous random variable, if you have a pdf f(x) on the interval a to b,

P(c≤x≤d)=∫dcf(x)dx
E(X)=μ=∫baxf(x)dx
Var(X)=∫ba(x−μ)2f(x)dx

For this problem, you are given with pdf f(x) and the interval is given. So you need to evaluate the integral.

The Attempt at a Solution

∫baf(x)dx= ∫8,2((x-2)/18))... I honestly lose it from here.

Yes, your notes are a mess. Let's see if we can straighten things out.

For this problem, the expectation E(x) and the Variance Var (x) are not required.

You are given a certain probability density function (pdf) and asked to find the probability that x will lie between 6 and 8.

The cumulative density function (cdf) is the integral of the pdf from -∞ to +∞ and by definition:

\int^{∞}_{-∞}[pdf (x)] dx = 1

This is not a difficult integral to evaluate since your pdf = 0 except between x = 2 and x = 8.
You should check your pdf to make sure that the relation above is satisfied.

Since the area of the pdf over the entire range of x is 1, this means that all possible outcomes of whatever the pdf represents have occurred. If you want to find the probability of occurrence of whatever is represented by the pdf on a certain interval of x, you would take the integral of the pdf between the limits of that interval.

 

[billbennett770]

okay then. From P(6<x≤8)

∫ f(x) dx = ((x²/2) -2x)/18

at x = 6 : ((6²/2) -2*6)/18= (18 -12)/18 = 1/3

at x = 8 : ((8²/2 -2*8)) = ((32 -16))/18 = 16/18 = 8/9

: 1/3 < P(6<x≤8) ≤ 8/9

 

[Simon Bridge]

You should only have one number - not a range.

Have you had any lessons on how to evaluate a definite integral?
http://tutorial.math.lamar.edu/Classes/CalcI/ComputingDefiniteIntegrals.aspx

 

[billbennett770]

Not as well I thought I did, or at least not how it pertains to continuous proability density. First, I am really perplexed why I am given 2 ≤ x ≤ 8 --or is this just another way of saying [2, 8]? Also, how do you evaluate P(6 <x ≤8)? I have only done int he past when the question employs both ≤ and , as in P(6 ≤x ≤8).

For example, here is a previous question I did:

Is f(x) 2/(x^2) on [1,2] a probability density function? Why or why not? If it is, find P[1.5 ≤ x ≤ 2]

∫ba f(x)dx = 2,1 2/(x^2) dx = [-2/x] 2, 1 = -1+2=1

So yes because it is postive.


∫1,5,2 (-2/x) = 1/3

 

[billbennett770]

So it it just (8/9) - (1/3)

 

[Simon Bridge]

Not as well I thought I did, or at least not how it pertains to continuous proability density.

Exactly the same as it does to anything.

First, I am really perplexed why I am given 2 ≤ x ≤ 8

That just means that f(x) is defined on that interval.
What is $p(2\leq x\leq 8)$ - evaluate it.

--or is this just another way of saying [2, 8]?

Yes. $x\in (2,8]$ is the same as saying $2<x\leq 8$

Also, how do you evaluate P(6 <x ≤8)? I have only done int he past when the question employs both ≤ and , as in P(6 ≤x ≤8).

The probability is the area under f(x) inside the range. The difference in area inside a<x<b and inside a<x≤b is zero. Sketch it out for a bunch of arbitrary functions and see.

So it it just (8/9) - (1/3)

There you go.