If the probability density of $X$ is given by $$f(x)= \begin{cases} 630x^4(1-x)^4&& \text{for } 0 <x<1\\ 0 && \text{elsewhere.}\\ \end{cases}$$ Find the probability that it will take on a value within two standard deviations of the mean and compare this probability with the lower-bounded provided by Chebyshev's theorem.
Let $\sigma$ be the standard deviation and $\mu$ be the mean.
How does one find the variance? $\sigma^2$. I know one must use the formula $\sigma^2= \mu^{'}_{2}-\mu^{2}$
In order to solve this one must integrate.
= $\int 630x^4(1-x)^4dx$
I cannot go any further than this as I do not know how to find the upper and lowerbounds.
EDIT After doing some extensive math I figured out that
$\mu = .5$ because
$\mu = \int^{1}_{0} x\cdot630x^4(1-x)^4 dx = .5$
$\mu_{2} =\int_0^1 x^2\cdot630x^4(1-x)^4dx= \frac{3}{11}$
$\sigma^2= \mu_{2}-u^{2} = \frac{3}{11}-(\frac{1}{2})^2=.0227 \rightarrow \sqrt{.0227} = .20 \text{ or } .15$
So
$\int_{.20} 630x^4(1-x)^4$
Now in order to finish the problem we must find what the upperbound is and this where I am lost.
Chebyshev’s Theorem
If $\mu$ and $\sigma$ are the mean and the standard deviation of a random variable X, then for any positive constant k the probability is at least $1- \frac{1}{k^2}$ that X will take on a value within k standard deviations of the mean; symbolically
$$P(|x-\mu|<k \sigma) \ge 1- \frac{1}{k^2}, \sigma \neq 0$$
Proof
$$\sigma^2 = E[(X-\mu)^2] = \int^{\infty}_{-\infty} (x-\mu)^2f(x)dx$$
Diagram for Chebyshev’s theorem.
Then dividing the integral into three parts as shown above
$$\sigma^2 = \int^{\mu-k\sigma}_{-\infty} (x-\mu)^2f(x)dx+ \int^{\mu + k\sigma}_{\mu-k\sigma}(x-\mu)^2f(x)dx+\int^{\infty}_{\mu+k\sigma} (x-\mu)^2f(x)dx$$
Since the integrand $(x-\mu)^2f(x)$ is nonnegative we can from the inequality
$$\sigma^2 \ge \int^{\mu-k\sigma}_{-\infty} (x-\mu)^2f(x)dx+\int^{\infty}_{\mu+k \sigma} (x-\mu)^2f(x)dx $$
by deleting the second integral. Therefore since $(x-\mu)^2 \ge k^2\sigma^2 \text{ for } x \le \mu -k\sigma \text{ or } x \ge \mu+k\sigma \text{ it follows that }$
$$(\sigma)^2 \ge \int^{\mu-k\sigma}_{-\infty} k^2\sigma^2f(x)dx+\int^{\infty}_{\mu+k\sigma} k^2\sigma^2f(x)dx$$
and hence that
$$\frac{1}{k^2} \ge \int^{\mu-k\sigma}_{-\infty} f(x)dx+ \int^{\infty}_{\mu+k\sigma}f(x)dx$$
provide $\sigma^2 \neq 0$ Since the sum of the two integrals on the right-hand side is the probability that X will take on a value less than or equal to $\mu-k\sigma$ or greater than or equal to $\mu+k\sigma$, we have thus shown that
$$P(|X-\mu| \ge k\sigma) \le \frac{1}{k^2}$$
and it follows that
$$P(|X-\mu| \lt k\sigma) \ge 1 - \frac{1}{k^2}$$
$\endgroup$ 41 Answer
$\begingroup$You use the usual formulas. \begin{align*} E[X] &= \int_{-\infty}^\infty xf(x)\, = \int_{0}^1 xf(x)\,dx\\ E[X^2] &= \int_{-\infty}^\infty x^2 f(x)\,dx = \int_0^1x^2 f(x)\,dx \end{align*} where the second equality in both lines is true because the pdf is "$0$ elsewhere". I leave the integration to you. Finally $$\operatorname{Var}[X] = E[X^2]-(E[X])^2.$$
You found that $E[X] = \frac{1}{2} =\mu$, $E[X^2] = \frac{3}{11}$, and $\text{Var}[X] = \frac{1}{44} = \sigma^2$
Then the exercise asks for $P(|X-\mu|< 2\sigma)$. \begin{align*} P(|X-\mu|<2\sigma)&= P(-2\sigma<X-\mu<2\sigma)\\ &= P(-2\sigma+\mu<X<2\sigma+\mu)\\ &= \int_{-2\sigma+\mu}^{2\sigma+\mu} 630x^4(1-x)^4\,dx \tag 1\\ &= 0.962071 \end{align*} where $(1)$ is true by the definition of pdf. The theorem gives $$P(|X-\mu|<2\sigma) \geq 1-\frac{1}{2^2} = \frac{3}{4}.$$
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