IB Math AA HL Flashcards

Exam revision with strategy and commentary

Question 1 of 14 • 1a

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Question (5 marks)

**Question 1(a)** [5 marks]

A continuous random variable $X$ has a probability density function $f$ given by

$f(x) = \begin{cases} \frac{x^2}{a} + \frac{1}{b}, & 0 \leq x \leq 3, \text{ where } a,b \in \mathbb{Z}^+ \\ 0, & \text{otherwise} \end{cases}$

It is given that $P(X \leq 2) = 0.31481$ , correct to five decimal places.

Show that $a = 27$ and $b = 12$ , correct to the nearest integer.

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Solution

We use the normalization condition for a PDF:

$\int_{-\infty}^{\infty} f(x) \, dx = 1$

$\int_0^3 \left(\frac{x^2}{a} + \frac{1}{b}\right) dx = 1$

$\left[\frac{x^3}{3a} + \frac{x}{b}\right]_0^3 = 1$

$\frac{27}{3a} + \frac{3}{b} = 1$

$\frac{9}{a} + \frac{3}{b} = 1 \quad \text{...(1)}$

Using the given condition $P(X \leq 2) = 0.31481$ :

$\int_0^2 \left(\frac{x^2}{a} + \frac{1}{b}\right) dx = 0.31481$

$\left[\frac{x^3}{3a} + \frac{x}{b}\right]_0^2 = 0.31481$

$\frac{8}{3a} + \frac{2}{b} = 0.31481 \quad \text{...(2)}$

Solving equations (1) and (2) simultaneously using GDC:

$\boxed{a = 27 \text{ and } b = 12}$

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