Integrating factor question?

a. For what value of $\setminus \lambda$ is the following ( $\setminus {\int}_{1}^{2} \setminus l a m \mathrm{da} \setminus \sqrt[3]{{x}^{4}} \mathrm{dx}$ ) a density function? b. Find $\setminus \mu$ Also: does anyone know where I can find practice questions on this type of problem?

Apr 25, 2018

See below

Explanation:

What you have here is a "wannabe" probability density function (PDF):

• $m a t h b f P \left(x\right) = \setminus \lambda \setminus \sqrt[3]{{x}^{4}}$, for $1 \le x \le 2$.

But first you want to normalise the PDF, ie make this become true:

• ${\int}_{1}^{2} m a t h b f P \left(x\right) = 1$.

....which is the continuous analogue of the requirement of a discrete distribution:

• ${\sum}_{i} P \left({x}_{i}\right) = 1$

$\lambda$ is your normalisation constant .

So:

$\setminus {\int}_{1}^{2} \setminus \lambda \setminus \sqrt[3]{{x}^{4}} \setminus \mathrm{dx} = \lambda \setminus {\left(\frac{3 {x}^{\frac{7}{3}}}{7}\right)}_{1}^{2}$

$= \lambda \cdot \frac{3}{7} \left(4 \sqrt[3]{2} - 1\right) \textcolor{red}{= 1}$

$\implies \lambda = \frac{7}{3 \left(4 \sqrt[3]{2} - 1\right)} \approx 0.58 \approx \frac{7}{12}$

$m a t h \boldsymbol{P} \left(x\right) = \frac{7}{3 \left(4 \sqrt[3]{2} - 1\right)} \setminus \sqrt[3]{{x}^{4}}$

The mean $\mu$ is given by:

• $\mu = {\int}_{1}^{2} x \setminus m a t h \boldsymbol{P} \left(x\right) \setminus \mathrm{dx}$

....which is the continuous analogue of the mean of a discrete distribution:

• $\mu = {\sum}_{i} P \left({x}_{i}\right) \setminus {x}_{i}$

Here:

$\mu = \lambda \setminus {\int}_{1}^{2} \setminus {x}^{\frac{7}{3}} \setminus \mathrm{dx}$

$= \lambda {\left(\frac{3 {x}^{\frac{10}{3}}}{10}\right)}_{1}^{2}$

$\frac{7}{3 \left(4 \sqrt[3]{2} - 1\right)} \cdot \frac{3}{10} \left(8 \sqrt[3]{2} - 1\right)$

$\implies \mu = \frac{7 \left(8 \sqrt[3]{2} - 1\right)}{10 \left(4 \sqrt[3]{2} - 1\right)} \approx 1.57$