random continuous variable

If you did not know already, "μ" is the greek letter "mu", and it stands for micro, or \(10^-6\) (10^-6).

The "η" is the greek letter "eta", and it stands for the greek letter "eta", and it stands for the "Dirichlet eta function".  In mathematics, in the area of analytic number theory, the Dirichlet eta function is defined by the following Dirichlet series, which converges for any complex number having real part > 0: 

\({\displaystyle \eta (s)=\sum _{n=1}^{\infty }{(-1)^{n-1} \over n^{s}}={\frac {1}{1^{s}}}-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\cdots }\)

...Except this is not analytic number theory, and "μ" simply represents tthe mean of X and "η" simply represents the median of X. Questions b) and c) tell us that "μ"  = 1 and "η" = \(\dfrac{13}{12}\). (The top part was a joke...)

\(3\cdot μ = 3\) and \(3 - \dfrac{13}{12} = 1\dfrac{11}{12}\). Question d) asks us to find the probabliltiy that \(X\lt 1\dfrac{11}{12}\). \(X\) can only be greater than \(1\dfrac{11}{12}\) when \(f(x) = \dfrac{3-x}{4}\). I believe you can go on from here and find the probabililty that \(X\lt 1\dfrac{11}{12}\). If you need more help, you could re-state the question or ask for more information in forums. 

- PartialMathematician

what are you going on about partialmathematician?

\(\text{for (d) you've already calculated }\mu=\dfrac{13}{12} \text{ and }\eta=1\\ \text{so you are asked to find }P\left[X < 3 \cdot \dfrac{13}{12} - 1\right] =\\ P\left[X <\dfrac 9 4\right]\)

\(\text{this will be}\\ \displaystyle \int \limits_0^1~\dfrac 1 2~dx + \int_1^{\frac 9 4}\dfrac{3-x}{4}~dx = \dfrac{119}{128}\\ \)