Let f(x)=30 x^2(1-x)^2 for 0⩽x ⩽1 and f(x)=0 for all other values of x . (a) Verify that f

Let f(x)=30 x^2(1-x)^2 for 0⩽x ⩽1 and f(x)=0 for all other...

Let $f(x)=30 x^{2}(1-x)^{2}$ for 0$\leqslant x \leqslant 1$ and $f(x)=0$ for all other values of $x .$ \begin{equation} \begin{array}{l}{\text { (a) Verify that } f \text { is a probability density function. }} \\ {\text { (b) Find } P\left(X \leqslant \frac{1}{3}\right)}\end{array} \end{equation}

Let $f(x)=30 x^{2}(1-x)^{2}$ for 0$\leqslant x \leqslant 1$ and $f(x)=0$ for all
other values of $x .$
\begin{equation}
\begin{array}{l}{\text { (a) Verify that } f \text { is a probability density function. }} \\ {\text { (b) Find } P\left(X \leqslant \frac{1}{3}\right)}\end{array}
\end{equation}

Key Concepts

Probability Density Function

A probability density function (pdf) describes the relative likelihood of a continuous random variable taking on a specific value. For a function to be a valid pdf, it must be non-negative for all possible values and integrate over its entire domain to equal 1. This normalization condition ensures that the total probability of all outcomes is exactly one.

Cumulative Distribution Function and Integration

The cumulative distribution function (CDF) for a continuous random variable is computed by integrating the pdf from the lower bound of the variable’s domain up to a given value. The CDF represents the probability that the variable is less than or equal to that value, thus linking the integral of the density function to real-world probabilities.

Transcript

0:00 All right.

00:01 To check that f is a probability density function, we need to check that f of x is non -negative everywhere, and that it integrates to one from negative infinity to infinity.

00:16 So first of all, f of x is either equal to zero or 30 x squared 1 minus x squared.

00:34 Yeah, i got that right.

00:36 Okay, so zero, first of all, is clearly greater than are equal to zero.

00:43 30 is non -negative, x squared squared, squares are non -negative, and one minus x squared also non -negative.

00:51 So this entire expression is non -negative.

00:55 So this first condition is good.

00:59 Next condition, since f of x is equal to zero outside of the interval from 0 to 1, we only need to look at the interval from 0 to 1, where f of x equals 30 x squared times 1 minus x squared dx.

01:23 We need to compute this and make sure that it's 1.

01:27 Well, this becomes the integral from 0 to 1 of 30 times, uh, hang on, of 30 x squared times 1 minus x squared is 1 minus 2x plus x squared the x this constant of 30 i'm going to go ahead and factor out to the front so we get 30 integral from 0 to 1 multiplying the x squared out we get x squared minus 2x cubed plus x to the 4th dx and the reason we've multiplied this whole expression out is now we can use the power rule so we get 30 times the antiderivative here is x cubed over 3 minus 2x to the 4th over 4 plus x to the 5th over 5.

02:37 And this is evaluated from 0 to 1.

02:41 Now, when we plug in zero, this entire expression becomes zero, so we only need to plug in one.

02:50 So what we are looking at is 30 times one -third minus one -half plus one -fifth.

03:05 Excuse me.

03:06 So this is 30 times 10 minus 15 plus 6...

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