What is the cdf of the area of a square with a uniform distribution over (0, 5)?

[theloathedone]

Homework Statement

Let the random variable X represent the length of the side of a square. It has a uniform distribution over the interval (0, 5).

What is the cumulative distribution function for the area of the square, Y?

Homework Equations

F(x) = 0.2x (the cdf of the side).

The Attempt at a Solution

So I tried simply squaring F(x), giving 0.04x^2, which is incorrect since F(25) = 25 instead of 1. Also, it wouldn't make sense for the probability of the largest areas to have the highest probability, since:
P(4.9 < X < 5.0) = 0.02
Therefore P(4.9^2 < X < 5.0^2) = 0.02
But going by 0.04x^2 we get 0.08 or something.

Also, I couldn't find anything like this in my textbook (I'm a high school student), is there any good website to line this stuff from?

Thanks in advance.

 

[VietDao29]

Homework Statement

Let the random variable X represent the length of the side of a square. It has a uniform distribution over the interval (0, 5).

What is the cumulative distribution function for the area of the square, Y?

Homework Equations

F(x) = 0.2x (the cdf of the side).

The Attempt at a Solution

So I tried simply squaring F(x), giving 0.04x^2, which is incorrect since F(25) = 25 instead of 1. Also, it wouldn't make sense for the probability of the largest areas to have the highest probability, since:
P(4.9 < X < 5.0) = 0.02
Therefore P(4.9^2 < X < 5.0^2) = 0.02
But going by 0.04x^2 we get 0.08 or something.

Also, I couldn't find anything like this in my textbook (I'm a high school student), is there any good website to line this stuff from?

Thanks in advance.

No, why are you squaring F(x)?

To handle this type of problem, we often try to find the cdf of S first, then (if the problem asks further for pdf function) we can obtain it by differentiating the cdf of S.

So, it goes like this:
\begin{align*}F_S(x) = P(S \le x) = P(X^2 \le x) &amp;= \left\{ \begin{array}{ll} 0 &amp; , \mbox{ if } x \le 0 \\ P(-\sqrt{x} \le X \le \sqrt{x}) &amp; , \mbox{ if } x &gt; 0 \end{array} \right. \\<br /> &amp;= \left\{ \begin{array}{ll} 0 &amp; , \mbox{ if } x \le 0 \\ P(-\sqrt{x} \le X \le 0) + P(0 &lt; X \le \sqrt{x}) &amp; , \mbox{ if } x &gt; 0 \end{array} \right. \\<br /> &amp;= ...<br /> \end{align}

Can you go from here, :)

 

[HallsofIvy]

You are going the wrong way. In order that x^2&lt; a, we must have x&lt;\sqrt{a}. Take the square root of 0.2x, not the square!

 

[theloathedone]

Ah I see where I went wrong.

But it should be 0.2\sqrt{x} and not \sqrt{0.2x} right?

Thanks for the speedy responses!

 

[VietDao29]

Ah I see where I went wrong.

But it should be 0.2\sqrt{x} and not \sqrt{0.2x} right?

Thanks for the speedy responses!

Yup, that's correct. Congratulations. :)

But, remember that F_S(x) = 0.2\sqrt{x} on the interval (0; 25), it takes other values elsewhere.

 

[theloathedone]

yup. it's 0 for x < 0 and 1 for x > 25. Thanks for the help :)