Math / Data & Statistics

QuestionLet $X$ be a random variable with density $f(x)=\frac{2 x}{8^{2}} \cdot 1(0,8)(x)$ . Find the quantile of rank $\frac{1}{15}$ for this variable:
Answer. $\square$
For random variable $\backslash(X V)$ as above, find the expected value of $\backslash(8.9 X+8.9 \backslash)$ .
Answer: $\square$

Studdy Solution

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 $q$ :
$q^2 = \frac{64}{15}$
$q = \sqrt{\frac{64}{15}}$
$q = \frac{8}{\sqrt{15}}$

STEP 6

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:
$\boxed{\frac{169.1}{3}}$

Was this helpful?

QuestionLet $X$ be a random variable with density $f(x)=\frac{2 x}{8^{2}} \cdot 1(0,8)(x)$ . Find the quantile of rank $\frac{1}{15}$ for this variable:
Answer. $\square$
For random variable $\backslash(X V)$ as above, find the expected value of $\backslash(8.9 X+8.9 \backslash)$ .
Answer: $\square$

Studdy Solution

STEP 1

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 $q$ :
$q^2 = \frac{64}{15}$
$q = \sqrt{\frac{64}{15}}$
$q = \frac{8}{\sqrt{15}}$

STEP 6

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Normalize the density function.2. Find the cumulative distribution function (CDF).3. Determine the quantile of rank 115\frac{1}{15}151​.4. Calculate the expected value of 8.9X+8.9 8.9X + 8.9 8.9X+8.9.

STEP 3

Normalize the density function f(x) f(x) f(x).The function is already normalized because:∫082x82 dx=1 \int_{0}^{8} \frac{2x}{8^2} \, dx = 1 ∫08​822x​dx=1

STEP 4

Find the cumulative distribution function (CDF) F(x) F(x) F(x).F(x)=∫0x2t64 dt=[t264]0x=x264 F(x) = \int_{0}^{x} \frac{2t}{64} \, dt = \left[ \frac{t^2}{64} \right]_{0}^{x} = \frac{x^2}{64} F(x)=∫0x​642t​dt=[64t2​]0x​=64x2​

STEP 5

Determine the quantile of rank 115\frac{1}{15}151​.Set F(q)=115 F(q) = \frac{1}{15} F(q)=151​:q264=115 \frac{q^2}{64} = \frac{1}{15} 64q2​=151​Solve for q q q:q2=6415 q^2 = \frac{64}{15} q2=1564​q=6415 q = \sqrt{\frac{64}{15}} q=1564​​q=815 q = \frac{8}{\sqrt{15}} q=15​8​

STEP 6

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

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 $8.9X + 8.9$ .

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

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

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