Suppose that Y has density function f(y) = ky(1 − y) if 0 ≤ y ≤ 1 0, elsewhere
A) Find the value of k that makes f (y) a probability density function.
B) Find P(Y ≤ .4|Y ≤ .8).
C) Find the .95-quantile, i. e., find a point φ.95 such that P(Y ≤ φ.95) = .95.

Using probability concepts, it is found that

a) The value is k = 6.

b) P(Y ≤ .4|Y ≤ .8) = 0.554.

c) The 95th percentile is x = 0.86465.

The density function is:

Item a:

The condition that makes it a probability density function is:

Thus:

Then

The value is k = 6.

Item b:

This probability is:

Then

Thus, P(Y ≤ .4|Y ≤ .8) = 0.554.

Item c:

This is x for which:

Thus:

Solving a cubic equation with the help of a calculator, and considering , this is x = 0.86465.

A similar problem is given at link

I'm assuming

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

Compute the integral:

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

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

We have

which reduces to the cubic

3φ² - 2φ³ = 0.95

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

the height of the surface increases from the center out to the sides of the road.

step-by-step explanation:

as we move out we move away from the x-axis.

and as mentioned in the problem statement that height of a road increases when we move away from the center x.

hence proved that 3rd option

the height of the surface increases from the center out to the sides of the road.

is the correct option.