Problem 22 - 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, which is option (a).

Step 1: Calculate the Mean

The mean of the distribution is found by multiplying each value of \(X\) by its probability and summing the results: \( \mu = 2 \times 0.3 + 3 \times 0.4 + 4 \times 0.3 = 3 \). Thus, the mean \( \mu \) is 3.

Step 2: Calculate the Variance

Variance is calculated using the formula \( \sigma^2 = \sum (x_i - \mu)^2 P(x_i) \). Substitute the values: - \((2-3)^2 \times 0.3 = 1 \times 0.3 = 0.3\), - \((3-3)^2 \times 0.4 = 0 \times 0.4 = 0\), - \((4-3)^2 \times 0.3 = 1 \times 0.3 = 0.3\). Add these values: \( 0.3 + 0 + 0.3 = 0.6 \). Hence, the variance is 0.6.

Key Concepts

Understanding Probability Distribution

A probability distribution is a comprehensive way to describe a random variable and its possible outcomes. It consists of all the values that the random variable can take and their associated probabilities. When dealing with probability distributions, it's essential to ensure that the sum of all probabilities equals 1. This is because, in any scenario, the random variable must take some value from the given set.
For instance, in our exercise, the random variable \(X\) can take the values 2, 3, and 4, with corresponding probabilities of 0.3, 0.4, and 0.3 respectively. If you add up these probabilities, 0.3 + 0.4 + 0.3, you will see they sum up to 1.

• Probability distributions can be used to predict outcomes and understand the chances of different events happening.
• Both discrete and continuous variables can have probability distributions, but the exercise here focuses on the discrete kind.

These distributions help to set the groundwork for further statistical analysis, such as finding the mean and the variance.

Mean of Random Variables

The mean, also known as the expected value, of a random variable provides a central tendency or an average of all its possible outcomes. To calculate the mean of a discrete random variable, you multiply each outcome by its probability, then sum these results. This gives you a weighted average.

Using the exercise's distribution, the mean of \(X\) is calculated as follows:

• Multiply each value of \(X\) by its corresponding probability:
• \(2 \times 0.3 = 0.6\)
• \(3 \times 0.4 = 1.2\)
• \(4 \times 0.3 = 1.2\)

Add these values to get the mean: \(0.6 + 1.2 + 1.2 = 3\). Hence, the mean \(\mu\) is 3. This tells us that, on average, if you observe the random variable \(X\), its value is most likely to be around 3. The mean helps in laying down a benchmark to compare the other statistical measures, like variance.

Variance Calculation

Variance is a crucial measure that quantifies the degree of spread in a set of random variables from their mean. Calculating variance helps us understand how varied and dispersed the values are, allowing insights into the reliability and risk associated with the data.

To find the variance of a discrete random variable, you use the formula \(\sigma^2 = \sum (x_i - \mu)^2 P(x_i)\). Follow these steps as demonstrated in the exercise:

• For \(X = 2\): \((2 - 3)^2 \times 0.3 = 1 \times 0.3 = 0.3\)
• For \(X = 3\): \((3 - 3)^2 \times 0.4 = 0 \times 0.4 = 0\)
• For \(X = 4\): \((4 - 3)^2 \times 0.3 = 1 \times 0.3 = 0.3\)

Add all these calculated values together to get the variance: \(0.3 + 0 + 0.3 = 0.6\).
The variance in this case is 0.6, indicating that the values of \(X\) are somewhat spread around the mean. A lower variance implies the data points are closer to the mean, whereas a higher variance indicates a wider spread.