# Math Help - Probability Distribution for a continuous random variable

1. ## Probability Distribution for a continuous random variable

The length of time required by students to complete a exam is a random variable with a density function given by:

f(x) = cx^3 + x if 0 <= x <= 1
0 otherwise

1) find c
2) Find the distribution function F(y).
3) Find the probability that a randomly selected student will finish the exam in less than half an hour
4) Given that a particular student needs at least 0.25 hours to complete the exam. What is the probability Heather will require at least 0.5 hours to complete the exam?

This is what I did:
1) I find the integral from 1 to 0 of x^3 + x
c[(x^4)/4 + (x^2)/2] = 1
c[(1/4) + (1/2) = 1
c(3/4) = 1
c is 4/3.

2)
4/3[(x^4)/4 + (x^2)/2]
[(x^4)/3] + [(2x^2)/3]

Is that correct? What about the rest of the problem? The book doesn't have any example so I am stuck.

2. Originally Posted by dkmathguy
The length of time required by students to complete a exam is a random variable with a density function given by:

f(x) = cx^3 + x if 0 <= x <= 1
0 otherwise

1) find c
2) Find the distribution function F(y).
3) Find the probability that a randomly selected student will finish the exam in less than half an hour
4) Given that a particular student needs at least 0.25 hours to complete the exam. What is the probability Heather will require at least 0.5 hours to complete the exam?

This is what I did:
1) I find the integral from 1 to 0 of x^3 + x
c[(x^4)/4 + (x^2)/2] = 1
c[(1/4) + (1/2) = 1
c(3/4) = 1
c is 4/3.

2)
4/3[(x^4)/4 + (x^2)/2]
[(x^4)/3] + [(2x^2)/3]

Is that correct? What about the rest of the problem? The book doesn't have any example so I am stuck.
Your answer for (1) is incorrect: c only multiplies the first term (you have corrected this in post #3). Your answer for (2) is incomplete (a common error made by students) and therefore wrong. The correct answer is:

$F(x) = 0$ for $x < 0$
$F(x) = \frac{x^4}{2} + \frac{x^2}{2}$ for $0 \leq x \leq 1$ (from post #3).
$F(x) = 1$ for $x > 1$.

You probably think that I'm being pedantic here, but I'm not.

(3) By definition: Pr(X < 1/2) = F(1/2) = ....

(4) Conditional probability is required: $\Pr(X > 0.5 | X > 0.25) = \frac{\Pr(X > 0.5)}{\Pr(X > 0.25)}$ (from Bayes Theorem).

3. Wait, I think I mess up on 1. The correct question is f(x) = cx^3 + x, not f(x) = c(x^3 + x)

So is this correct?

c((x^4)/4) + (x^2)/2 = 1
c((1/4)) + (1/2) = 1
c(1/4) = 1/2
c is 2

For 2, I have to right all 3 of those for the answer?

4. Originally Posted by dkmathguy
Wait, I think I mess up on 1. The correct question is f(x) = cx^3 + x, not f(x) = c(x^3 + x)

So is this correct?

c((x^4)/4) + (x^2)/2 = 1
c((1/4)) + (1/2) = 1
c(1/4) = 1/2
c is 2

[snip]
Yes.

5. (3) By definition: Pr(X < 1/2) = F(1/2) = ....

((.5)^4)/2 + ((.5)^2)/2 = .15625

(4) Conditional probability is required: (from Bayes Theorem).

((.5)^4)/2 + ((.5)^2)/2 = 0.15625
((.25)^4)/2 + ((.25)^2)/2 = 0.03320

(1 - 0.15625)/(1 - 0.03320) = .8727.

Is that correct? I am so tired, its 3 a.m over here. Good night, thank Mr. F, I learn a lot today. Thank you very much.