Math Snap

STEP 1

1. The random variable $X$ has a probability density function (pdf) given by $f(x) = \frac{2x}{8^2} \cdot 1_{(0,8)}(x)$.
2. The quantile of rank $\frac{1}{15}$ is the value $q$ such that $P(X \leq q) = \frac{1}{15}$.
3. The expected value of a linear transformation $aX + b$ is given by $aE(X) + b$.

STEP 2

1. Normalize the density function.
2. Find the cumulative distribution function (CDF).
3. Determine the quantile of rank $\frac{1}{15}$.
4. Calculate the expected value of $8.9X + 8.9$.

STEP 3

Normalize the density function $f(x)$.
The function is already normalized because:
$$\int_{0}^{8} \frac{2x}{8^2} \, dx = 1$$

STEP 4

Find the cumulative distribution function (CDF) $F(x)$.
$$F(x) = \int_{0}^{x} \frac{2t}{64} \, dt = \left[ \frac{t^2}{64} \right]_{0}^{x} = \frac{x^2}{64}$$

STEP 5

Determine the quantile of rank $\frac{1}{15}$.
Set $F(q) = \frac{1}{15}$:
$$\frac{q^2}{64} = \frac{1}{15}$$ Solve for $q$:
$$q^2 = \frac{64}{15}$$ $$q = \sqrt{\frac{64}{15}}$$ $$q = \frac{8}{\sqrt{15}}$$

Calculate the expected value of $X$.
$$E(X) = \int_{0}^{8} x \cdot \frac{2x}{64} \, dx = \int_{0}^{8} \frac{2x^2}{64} \, dx = \left[ \frac{2x^3}{192} \right]_{0}^{8}$$ $$E(X) = \frac{2 \times 512}{192} = \frac{1024}{192} = \frac{16}{3}$$ Calculate the expected value of $8.9X + 8.9$:
$$E(8.9X + 8.9) = 8.9E(X) + 8.9$$ $$E(8.9X + 8.9) = 8.9 \times \frac{16}{3} + 8.9$$ $$E(8.9X + 8.9) = \frac{142.4}{3} + 8.9$$ $$E(8.9X + 8.9) = \frac{142.4}{3} + \frac{26.7}{3}$$ $$E(8.9X + 8.9) = \frac{169.1}{3}$$ The quantile of rank $\frac{1}{15}$ is:
$$\boxed{\frac{8}{\sqrt{15}}}$$ The expected value of $8.9X + 8.9$ is:
$$\boxed{\frac{169.1}{3}}$$