I'm assuming
(a) <em>f(x)</em> is a valid probability density function if its integral over the support is 1:
Compute the integral:
So we have
<em>k</em> / 6 = 1 → <em>k</em> = 6
(b) By definition of conditional probability,
P(<em>Y</em> ≤ 0.4 | <em>Y</em> ≤ 0.8) = P(<em>Y</em> ≤ 0.4 and <em>Y</em> ≤ 0.8) / P(<em>Y</em> ≤ 0.8)
P(<em>Y</em> ≤ 0.4 | <em>Y</em> ≤ 0.8) = P(<em>Y</em> ≤ 0.4) / P(<em>Y</em> ≤ 0.8)
It makes sense to derive the cumulative distribution function (CDF) for the rest of the problem, since <em>F(y)</em> = P(<em>Y</em> ≤ <em>y</em>).
We have
Then
P(<em>Y</em> ≤ 0.4) = <em>F</em> (0.4) = 0.352
P(<em>Y</em> ≤ 0.8) = <em>F</em> (0.8) = 0.896
and so
P(<em>Y</em> ≤ 0.4 | <em>Y</em> ≤ 0.8) = 0.352 / 0.896 ≈ 0.393
(c) The 0.95 quantile is the value <em>φ</em> such that
P(<em>Y</em> ≤ <em>φ</em>) = 0.95
In terms of the integral definition of the CDF, we have solve for <em>φ</em> such that
We have
which reduces to the cubic
3<em>φ</em>² - 2<em>φ</em>³ = 0.95
Use a calculator to solve this and find that <em>φ</em> ≈ 0.865.