What is the domain of y = log5x?

Suppose that Y has density function

zvonat [6]

I'm assuming

$f(y)=\begin{cases}ky(1-y)&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(a) f(x) is a valid probability density function if its integral over the support is 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=k\int_0^1 y(1-y)\,\mathrm dy=k\int_0^1(y-y^2)\,\mathrm dy=1$

Compute the integral:

$\displaystyle\int_0^1(y-y^2)\,\mathrm dy=\left(\frac{y^2}2-\frac{y^3}3\right)\bigg|_0^1=\frac12-\frac13=\frac16$

So we have

k / 6 = 1 → k = 6

(b) By definition of conditional probability,

P(Y ≤ 0.4 | Y ≤ 0.8) = P(Y ≤ 0.4 and Y ≤ 0.8) / P(Y ≤ 0.8)

P(Y ≤ 0.4 | Y ≤ 0.8) = P(Y ≤ 0.4) / P(Y ≤ 0.8)

It makes sense to derive the cumulative distribution function (CDF) for the rest of the problem, since F(y) = P(Y ≤ y).

We have

$\displaystyle F(y)=\int_{-\infty}^y f(t)\,\mathrm dt=\int_0^y6t(1-t)\,\mathrm dt=\begin{cases}0&\text{for }y$

Then

P(Y ≤ 0.4) = F (0.4) = 0.352

P(Y ≤ 0.8) = F (0.8) = 0.896

and so

P(Y ≤ 0.4 | Y ≤ 0.8) = 0.352 / 0.896 ≈ 0.393

(c) The 0.95 quantile is the value φ such that

P(Y ≤ φ) = 0.95

In terms of the integral definition of the CDF, we have solve for φ such that

$\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=0.95$

We have

$\displaystyle\int_{-\infty}^\varphi f(y)\,\mathrm dy=\int_0^\varphi 6y(1-y)\,\mathrm dy=(3y^2-2y^3)\bigg|_0^\varphi = 0.95$

which reduces to the cubic

3φ² - 2φ³ = 0.95

Use a calculator to solve this and find that φ ≈ 0.865.

8 0

3 years ago

2 < 17 ÷ e please help I don't know how to solve this

IceJOKER [234]

I don't know either because I don't know what the e equals to

7 0

2 years ago

Solve each quadratic equation by completing the square. 6x2 – 28x = - 16

Nataly_w [17]

Answer:

<h3> There is no real solution</h3>

Step-by-step explanation:

$6x^2-28x = - 16\\\\6x^2-28x + 16=0\\\\3x^2-7x+8=0\\\\9x^2-21x+24=0\\\\(3x)^2-2\cdot3x\cdot\frac72+(\frac72)^2-(\frac72)^2+24=0\\\\(3x-\frac72)^2-\frac{49}4+24=0\\\\(3x-\frac72)^2-12\frac{1}4+24=0\\\\(3x-\frac72)^2+12\frac{3}4=0\\\\(3x-\frac72)^2=-12\frac{3}4$

4 0

3 years ago

The measures of the angles of a triangle are shown in the figure below. Solve for x.

andrey2020 [161]

x=180°-(102°+43°)=180°-145°=35°

7 0

2 years ago

Ana Participated in each charity walk she raise $.25 and each 1/2 That she walked the first day and I walked 11 miles a second d

Tresset [83]

Question:

Ana participated in a charity walk. She raised $0.25 for each 1/2 mile that she walked.The first day Ana walked 11 miles.The second day, she walked 14 miles.How much money did Ana raised?

Answer:

Ana raised $ 12.5

Solution:

From given question,

First day walk = 11 miles

Second day walk = 14 miles

Let us first calculate the total distance she walked

Total distance = first day walk + second day walk

Total distance = 11 + 14 = 25 miles

Thus she walked for 25 miles

Given that,

She raised $0.25 for each 1/2 mile that she walked

$\frac{1}{2} \text{ mile} = 0.25 \text{ dollars }$

Therefore, for 1 mile we get,

$\frac{1}{2} \times 2 \text{ mile} = 0.25 \times 2 \text{ dollars }\\\\1 \text{ mile } = 0.5 \text{ dollars }$

Now calculate for 25 miles

$25 \text{ mile } = 25 \times 0.5 = 12.5 \text{ dollars }$

Thus she raised $ 12.5

4 0

2 years ago