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.

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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.

$32,660

step-by-step explanation: