Problem 21 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A random variable \(X\) is specified by the following distribution law: \(\begin{array}{lccc}X: & 2 & 3 & 4 \\ P(X=x): & 0.3 & 0.4 & 0.3\end{array}\) Then the variance of this distribution is (a) \(0.6\) (b) \(0.7\) (c) \(0.77\) (d) \(1.55\) ution $$ \begin{aligned} &\text { (a) } \text { Mean }=(2)(0.3)+(3)(0.4)+(4)(0.3)=3 \\ &x^{2}=\operatorname{Variance}(x) \sum(x-\bar{x})^{2} P \\ &\quad=(2-3)^{2}(0.3)+(3.3)^{2}(0.4)+(4-3)^{2} \times 0.3 \\ &=0.6 \end{aligned} $$

Step-by-Step Solution

Answer

The variance of the distribution is 0.6, so the answer is (a) 0.6.

Step 1: Calculate the Expected Value (Mean)

The expected value \( \mu \) is calculated by multiplying each value of \( X \) by its corresponding probability and then summing all the products: \[ \mu = (2 \times 0.3) + (3 \times 0.4) + (4 \times 0.3) = 0.6 + 1.2 + 1.2 = 3 \]

Step 2: Calculate Squared Deviations from Mean

Calculate the squared deviations from the mean for each value of \( X \). The mean is 3. For \( X = 2 \), the deviation is \( (2 - 3)^2 = 1 \).For \( X = 3 \), the deviation is \( (3 - 3)^2 = 0 \).For \( X = 4 \), the deviation is \( (4 - 3)^2 = 1 \).

Step 3: Multiply Squared Deviations by Probabilities

Multiply each squared deviation by the corresponding probability:\[ (2-3)^2 \times 0.3 = 1 \times 0.3 = 0.3 \]\[ (3-3)^2 \times 0.4 = 0 \times 0.4 = 0 \]\[ (4-3)^2 \times 0.3 = 1 \times 0.3 = 0.3 \]

Step 4: Sum the Results to Find Variance

Add up the products from Step 3 to find the variance:\[ \text{Variance} = 0.3 + 0 + 0.3 = 0.6 \]

Step 5: Final Step: Conclude with the Variance

The variance of the distribution for the random variable \( X \) is \( 0.6 \). Thus, the answer is (a) \( 0.6 \).

Key Concepts

Discrete Probability Distribution

In probability theory, a discrete probability distribution is a key concept that helps us understand how probabilities are assigned to discrete random variables. Discrete random variables are those that can take on a finite or countably infinite number of distinct values. For each possible value of the random variable, the probability distribution describes the likelihood of that value occurring.

To put it simply, think of a discrete probability distribution as a list, or sometimes a table. It shows which values a random variable can have and how likely each of those values is. In our exercise, we have a discrete random variable, X, which can take on the values 2, 3, and 4. Each of these values is associated with probabilities 0.3, 0.4, and 0.3, respectively. These probabilities sum up to 1, as they should in any valid probability distribution.

Understanding a discrete probability distribution allows us to analyze and calculate various statistical measures about the random variable, such as its expected value and variance.

Expected Value (Mean)

The expected value, also known as the mean, is a measure of the central tendency of a random variable. In simple terms, it gives us a general idea of what value the random variable is likely to have on average.

The expected value is calculated by multiplying each possible value of the random variable by its probability and then summing up all these products. For our exercise, the expected value of X is calculated as follows: \[ \mu = (2 \times 0.3) + (3 \times 0.4) + (4 \times 0.3) = 0.6 + 1.2 + 1.2 = 3 \] In this example, the expected value (3) signifies that on average, if we were to repeat the process many times, the random variable X would tend to take on the value 3.

Remember, the expected value does not necessarily have to be one of the possible values the random variable can take. Instead, it serves as a tool that summarizes the distribution into one single, "average" number.

Squared Deviations

To measure how spread out the values of a random variable are around the mean, we look at the squared deviations. This involves calculating the difference between each possible value of the variable and the mean, then squaring this difference. Squaring ensures that all deviations are treated as positive contributions.

For our random variable X with a mean of 3, we calculate the squared deviations as:

• For X = 2: \((2 - 3)^2 = 1\)
• For X = 3: \((3 - 3)^2 = 0\)
• For X = 4: \((4 - 3)^2 = 1\)

The squared deviations alone are informative, but multiplying them by their corresponding probabilities gives a weighted average, leading us to the concept of variance.

This approach of considering squared deviations is core to variance since it quantifies how much the values differ from the expected value.

Probability Theory

Probability theory is the branch of mathematics concerned with analyzing random phenomena. At its core, it provides tools for understanding and modeling uncertain events. It is foundational in many fields such as statistics, finance, and various sciences.

In the realm of discrete random variables, probability theory provides us with methods to calculate and interpret important measures such as the expected value and variance. These measures help us comprehend and predict the behavior of random variables, as seen in exercises like ours.

Probability theory assures us that the sum of the probabilities in any discrete probability distribution will always be 1, as the occurrence of one of the possible outcomes is certain. It also assists in making the right calculations for variance, involving probabilities, squared deviations, and ultimately understanding distributions comprehensively. Understanding these principles is crucial for anyone dealing with data and randomness in real-world applications.