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Probability - Mean of a random variable

Grade 12CBSE

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

🔑Concepts

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A random variable XX is a real-valued function whose domain is the sample space of a random experiment. Visually, you can imagine this as a mapping where every outcome in the sample space (a set of points) is linked by an arrow to a specific numerical value on the real number line.

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A Discrete Random Variable is one which can take only a finite or countably infinite number of values. Its probability distribution is often visualized as a probability mass function (PMF) plot, where vertical bars are drawn at each possible value xix_i with heights corresponding to their probabilities pip_i.

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For any discrete probability distribution, the sum of all probabilities must equal 11, i.e., ∑pi=1\sum p_i = 1. In a tabular representation, the top row lists the values of XX and the bottom row lists the corresponding p(x)p(x); the total area represented by these probabilities (if viewed as a histogram) always sums to unity.

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The Mean of a random variable, also called the Expected Value E(X)E(X), is the weighted average of all possible values that the random variable can take. It represents the long-term average outcome if the experiment were repeated an infinite number of times.

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Visually, the mean can be understood as the 'center of mass' or 'balance point' of the distribution. If the probability distribution were placed on a see-saw (lever) with weights pip_i at positions xix_i, the mean μ\mu is the exact point where the see-saw would balance perfectly.

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The Mean does not necessarily have to be one of the possible values of the random variable. For example, the mean of a fair die roll is 3.53.5, even though a die can never actually show 3.53.5. This represents the central tendency of the numerical outcomes.

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Linearity of Expectation: The mean of a linear transformation of a random variable is the linear transformation of its mean. If Y=aX+bY = aX + b, then E(Y)=aE(X)+bE(Y) = aE(X) + b. This shifts the 'center' of the distribution graph along the x-axis and scales its spread.

📐Formulae

μ=E(X)=∑i=1nxipi\mu = E(X) = \sum_{i=1}^{n} x_i p_i

P(X=xi)=pi≥0P(X = x_i) = p_i \ge 0

∑i=1npi=1\sum_{i=1}^{n} p_i = 1

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

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

💡Examples

Problem 1:

A fair coin is tossed twice. Let XX be the number of heads obtained. Find the mean (Expected Value) of XX.

Solution:

  1. Identify the sample space: S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}.
  2. Define the values of XX: XX can be 00 (for TTTT), 11 (for HT,THHT, TH), or 22 (for HHHH).
  3. Calculate probabilities:
  • P(X=0)=14P(X=0) = \frac{1}{4}
  • P(X=1)=24=12P(X=1) = \frac{2}{4} = \frac{1}{2}
  • P(X=2)=14P(X=2) = \frac{1}{4}
  1. Apply the mean formula: E(X)=∑xipi=(0×14)+(1×12)+(2×14)E(X) = \sum x_i p_i = (0 \times \frac{1}{4}) + (1 \times \frac{1}{2}) + (2 \times \frac{1}{4}) E(X)=0+0.5+0.5=1E(X) = 0 + 0.5 + 0.5 = 1.

Explanation:

We first determine the probability distribution of the number of heads and then multiply each outcome by its probability. The result 11 indicates that on average, we expect one head per two tosses.

Problem 2:

The probability distribution of a random variable XX is given by P(X=x)=kxP(X=x) = kx for x=1,2,3x = 1, 2, 3 and P(X=x)=0P(X=x) = 0 otherwise. Find the value of kk and the mean of XX.

Solution:

  1. Find kk using ∑pi=1\sum p_i = 1: k(1)+k(2)+k(3)=1  ⟹  6k=1  ⟹  k=16k(1) + k(2) + k(3) = 1 \implies 6k = 1 \implies k = \frac{1}{6}.
  2. The distribution is: P(1)=16,P(2)=26,P(3)=36P(1)=\frac{1}{6}, P(2)=\frac{2}{6}, P(3)=\frac{3}{6}.
  3. Calculate the mean: 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:

First, we use the property that total probability equals 11 to find the unknown constant kk. Then, we use the standard mean formula to find the expected value.