IB Mathematics AI SL Concept of discrete random variables and their probability MAI Study Notes - New Syllabus
IB Mathematics AI SL Concept of discrete random variables and their probability MAI Study Notes
LEARNING OBJECTIVE
- Concept of discrete random variables and their probability distributions.
Key Concepts:
- Discrete random variables
- E (X) for discrete data.
- IBDP Maths AI SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
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- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 3
PROBABILITY DISTRIBUTION OF A RANDOM VARIABLE X
A random variable \( X \) takes on some values in a given domain at random!!! It may be A discrete variable takes on values in a finite or numerable set, while a continuous variable takes on values in some interval(s).
DISCRETE RANDOM VARIABLE
Let \( X \) be a variable which takes on the values
$ 10, 20, 30 $
with probabilities
$ 0.2, 0.3, 0.5 $
respectively. We often use a table:
$\rm x$ | $10$ | $20$ | $30$ |
$\text{P(X=x)}$ | $0.2$ | $0.3$ | $0.5$ |
1. All the probabilities are non-negative numbers; and
2. Their sum is always 1.
Then we say that \( X \) is a discrete random variable.
To express that the probability that \( X=10 \) is 0.2, we write:
$ P(X=10) = 0.2 $
Similarly, \( P(X=20) = 0.3 \) and \( P(X=30) = 0.5 \).
In general, for a discrete random variable \( X \) with:
$ \begin{array}{c|cccc}
X & x_1 & x_2 & x_3 & \cdots \\
\hline
P(X=x) & p_1 & p_2 & p_3 & \cdots \\
\end{array} $
it holds:
1. \( p_i \geq 0 \), for all \( i \).
2. \( \sum p_i = 1 \), i.e., \( p_1 + p_2 + p_3 + \cdots = 1 \).
We write:
$ P(X=x_1) = p_1, \quad P(X=x_2) = p_2, \quad \text{and so on.} $
(We also say that a probability function \( p: x_i \mapsto y_i \) is defined).
◆ THE EXPECTED VALUE \( \mu = E(X) \)
The mean \( \mu \) or otherwise the expected value \( E(X) \) is defined by:
$ E(X) = \sum x_i p_i = x_1 p_1 + x_2 p_2 + x_3 p_3 + \cdots $
For our example:
$ \begin{array}{c|ccc}
X & 10 & 20 & 30 \\
\hline
P(X=x) & 0.2 & 0.3 & 0.5 \\
\end{array} $
the expected value (otherwise the mean) is:
$ E(X) = 10 \times 0.2 + 20 \times 0.3 + 30 \times 0.5 = 23 $
NOTICE: Explanation for \( \mu = E(X) \)
In fact, the mean here is not different than the mean in statistics.
Consider the following ten numbers:
$ 10, 10, 20, 20, 20, 30, 30, 30, 30, 30 $
The probabilities to select 10, 20, or 30 are as in the table above.
The mean in statistics is also:
$ \mu = \frac{10 \times 2 + 20 \times 3 + 30 \times 5}{10} = 10 \times \frac{2}{10} + 20 \times \frac{3}{10} + 30 \times \frac{5}{10} = 23 $
Example Consider:
Given that \( E(X)=23 \), find the values of \( a \) and \( b \). ▶️Answer/ExplanationSolution: Solution We use two relations: The solution of the system is \( a = 0.2 \) and \( b = 0.3 \). The probability distribution applies in many betting games: |
Example Consider again the same table above. But now we select one of the numbers 10, 20, 30 at random. If we select 10, we earn 6 points. What is the expected number of points in one game? ▶️Answer/ExplanationSolution: Expected profit: $ \text{Expected profit} = 6 \times 0.2 + 1 \times 0.3 – 2 \times 0.5 = 0.5 $ That is, in each game we earn 0.5 points on average. |
◆ MEDIAN-MODE
These measures, known from statistics, are defined analogously:
MODE = The value \( X=a \) of the highest probability.
MEDIAN = The value \( X=m \) where the probability splits in two equal parts (0.5-0.5).
Example Consider again the probability distribution:
Find Mode and median. ▶️Answer/ExplanationSolution: Mode $P(X = 10) = 0.2$ The highest probability is 0.5, which occurs at $x = 30$, so: Median Cumulative probabilities The median is the smallest $x$ such that $P(X \leq x) \geq 0.5$. $\rm{\text{Median} = 20}$ |
◆ VARIANCE (Only for HL )
We define:
$ \text{Var}(X) = E(X-\mu)^2 $
that is:
$ \text{Var}(X) = (x_1-\mu)^2 \times p_1 + (x_2-\mu)^2 \times p_2 + (x_3-\mu)^2 \times p_3 + \cdots $
An equivalent definition is:
$ \text{Var}(X) = E(X^2) – \mu^2 $
where:
$ E(X^2) = x_1^2 \times p_1 + x_2^2 \times p_2 + x_3^2 \times p_3 + \cdots $
Example Consider again the probability distribution:
Using first and second both definition Find $ \text{Var}(X) $ ▶️Answer/ExplanationSolution: We have seen that \( \mu = E(X) = 23 \). 1. Using the first definition: 2. Using the second definition: |