Triangular probability density function: peak height from total probability and peak position from the mean

Triangular probability density function: peak height from total probability and peak position from the mean

IB Year 6 | Grade 12
6 marks

The continuous random variable \(X\) has probability density function \(\mathrm{f}\) given by \(\mathrm{f}(x)=\begin{cases}\dfrac{b}{a}x, & 0\leq x\leq a,\\ \dfrac{b}{a-8}(x-8), & a<x\leq8,\\ 0, & \text{otherwise},\end{cases}\) where \(0<a<8\) and \(b\in\mathbb{Q}^+\).

The diagram shows the graph of \(y=\mathrm{f}(x)\).

  1. Show that \(b=0.25\).[2]
  2. The mean of \(X\) is \(3.5\).
    Find the value of \(a\).[4]

Solution:

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Answer:(a) \(b=0.25\) (shown) (b) \(a=2.5\)

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