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Probability - Mean and Variance of a Random Variable

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Discrete Random Variable: A random variable XX is a real-valued function whose domain is the sample space of a random experiment. For a discrete variable, the values can be listed as a sequence x1,x2,...x_1, x_2, ... which, when plotted on a horizontal axis, appear as distinct, separated points.

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Probability Distribution Function (PDF): A mapping that assigns a probability pip_i to each possible value xix_i. Visually, this is represented by a probability histogram where each outcome xix_i has a bar of height pip_i, and the total area or sum of the heights of all bars must equal 11.

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Mean or Expected Value E(X)E(X): This is the weighted average of all possible values of XX. In a physical sense, if the probabilities were masses placed at positions xix_i on a beam, the mean would be the 'center of gravity' or the exact point where the beam would balance.

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Variance Var(X)Var(X): A statistical measure that describes the dispersion of the random variable values around the mean. On a graph, a 'flat' or 'wide' distribution indicates a high variance, while a 'tall' and 'narrow' distribution indicates the values are clustered closely around the mean.

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Standard Deviation ΃\sigma: This is the positive square root of the variance. It is often more intuitive than variance because it is expressed in the same units as the random variable XX, representing the typical distance an outcome is from the mean.

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Linearity of Expectation: A concept stating that for constants aa and bb, E(aX+b)=aE(X)+bE(aX + b) = aE(X) + b. Visually, adding bb shifts the entire probability distribution graph to the right, while multiplying by aa stretches or compresses the distribution.

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Variance of Linear Transformations: The variance of a transformed variable is given by Var(aX+b)=a2Var(X)Var(aX + b) = a^2 Var(X). This highlights that shifting a distribution (adding bb) does not change its spread, but scaling it (multiplying by aa) increases the spread by the square of the scale factor.

📐Formulae

∑i=1nP(xi)=1\sum_{i=1}^{n} P(x_i) = 1

Mean \mu = E(X) = \sum_{i=1}^{n} x_i p_i$

E(X2)=∑i=1nxi2piE(X^2) = \sum_{i=1}^{n} x_i^2 p_i

Variance \sigma^2 = Var(X) = E(X^2) - [E(X)]^2$

StandardDeviation΃=Var(X)Standard Deviation \sigma = \sqrt{Var(X)}

E(aX+b)=aE(X)+bE(aX + b) = aE(X) + b

Var(aX+b)=a2Var(X)Var(aX + b) = a^2 Var(X)

💡Examples

Problem 1:

A random variable XX has the following probability distribution: X={0,1,2,3}X = \{0, 1, 2, 3\} with P(X=0)=0.1,P(X=1)=0.4,P(X=2)=0.3,P(X=3)=0.2P(X=0) = 0.1, P(X=1) = 0.4, P(X=2) = 0.3, P(X=3) = 0.2. Calculate the Mean and Variance.

Solution:

  1. Calculate Mean E(X)E(X): E(X)=(0×0.1)+(1×0.4)+(2×0.3)+(3×0.2)E(X) = (0 \times 0.1) + (1 \times 0.4) + (2 \times 0.3) + (3 \times 0.2) E(X)=0+0.4+0.6+0.6=1.6E(X) = 0 + 0.4 + 0.6 + 0.6 = 1.6
  2. Calculate E(X2)E(X^2): E(X2)=(02×0.1)+(12×0.4)+(22×0.3)+(32×0.2)E(X^2) = (0^2 \times 0.1) + (1^2 \times 0.4) + (2^2 \times 0.3) + (3^2 \times 0.2) E(X2)=0+0.4+1.2+1.8=3.4E(X^2) = 0 + 0.4 + 1.2 + 1.8 = 3.4
  3. Calculate Variance Var(X)Var(X): Var(X)=E(X2)−[E(X)]2Var(X) = E(X^2) - [E(X)]^2 Var(X)=3.4−(1.6)2=3.4−2.56=0.84Var(X) = 3.4 - (1.6)^2 = 3.4 - 2.56 = 0.84

Explanation:

We first find the weighted average (Mean) by multiplying each value by its probability. Then, we find the average of the squares (E(X2)E(X^2)) to apply the shortcut formula for variance.

Problem 2:

Find the value of kk for the probability distribution P(X=x)=kxP(X=x) = kx for x=1,2,3x = 1, 2, 3. Then find the Mean.

Solution:

  1. Find kk: Since ∑P(X=x)=1\sum P(X=x) = 1, we have: k(1)+k(2)+k(3)=1k(1) + k(2) + k(3) = 1 6k=1  ⟹  k=166k = 1 \implies k = \frac{1}{6}
  2. Write the distribution: P(X=1)=16,P(X=2)=26,P(X=3)=36P(X=1) = \frac{1}{6}, P(X=2) = \frac{2}{6}, P(X=3) = \frac{3}{6}
  3. Calculate Mean E(X)E(X): E(X)=(1×16)+(2×26)+(3×36)E(X) = (1 \times \frac{1}{6}) + (2 \times \frac{2}{6}) + (3 \times \frac{3}{6}) E(X)=16+46+96=146=73≈2.33E(X) = \frac{1}{6} + \frac{4}{6} + \frac{9}{6} = \frac{14}{6} = \frac{7}{3} \approx 2.33

Explanation:

We use the property that total probability must sum to 11 to solve for the unknown constant kk, and then proceed with the standard mean formula.