Understanding the Problem: Variance of a Random Variable

The question asks for the variance of a random variable Y, given its relationship with another random variable X, where X follows a specific distribution.

We are given two random variables X and Y such that the sum of their values is always 100:

\( X + Y = 100 \)

This relationship allows us to express Y in terms of X:

\( Y = 100 - X \)

We are also told that X follows a Binomial distribution with parameters n = 100 and p = \(\frac{4}{5}\). The notation for this is \( X \sim B(n=100, p=\frac{4}{5}) \).

Calculating Variance using Variance Properties

To find the variance of Y, we can use the properties of variance. A key property is that for any constants 'a' and 'b' and a random variable X, the variance of aX + b is given by:

\( \text{Var}(aX + b) = a^2 \text{Var}(X) \)

In our case, Y = 100 - X. We can write this as \( Y = (-1)X + 100 \). Here, \( a = -1 \) and \( b = 100 \).

Using the variance property, we have:

\( \text{Var}(Y) = \text{Var}(100 - X) \)

\( \text{Var}(Y) = \text{Var}((-1)X + 100) \)

\( \text{Var}(Y) = (-1)^2 \text{Var}(X) \)

\( \text{Var}(Y) = 1 \cdot \text{Var}(X) \)

\( \text{Var}(Y) = \text{Var}(X) \)

So, the variance of Y is equal to the variance of X. Now we need to calculate the variance of X.

Finding the Variance of X (Binomial Distribution)

Since X follows a Binomial distribution with parameters n and p, the variance of X is given by the formula:

\( \text{Var}(X) = n \cdot p \cdot (1-p) \)

From the question, we know:

• Number of trials, \( n = 100 \)
• Probability of success, \( p = \frac{4}{5} \)

First, we need to calculate \( 1-p \), which is the probability of failure:

\( 1 - p = 1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5} \)

Now, substitute the values of n, p, and (1-p) into the variance formula for the Binomial distribution:

\( \text{Var}(X) = 100 \cdot \frac{4}{5} \cdot \frac{1}{5} \)

Perform the multiplication:

\( \text{Var}(X) = 100 \cdot \left(\frac{4 \cdot 1}{5 \cdot 5}\right) \)

\( \text{Var}(X) = 100 \cdot \frac{4}{25} \)

Now, calculate the final value:

\( \text{Var}(X) = \frac{100}{25} \cdot 4 \)

\( \text{Var}(X) = 4 \cdot 4 \)

\( \text{Var}(X) = 16 \)

Conclusion: Variance of Y

We found that \( \text{Var}(Y) = \text{Var}(X) \). Since \( \text{Var}(X) = 16 \), the variance of Y is also 16.

\( \text{Var}(Y) = 16 \)

Revision Table: Key Concepts

Additional Information: Understanding Variance and Binomial Distribution

Variance: Variance is a measure of how spread out a set of data or a probability distribution is. A high variance indicates that the data points are widely spread from the mean, while a low variance indicates that the data points are clustered closely around the mean. It is the expected value of the squared deviation from the mean, i.e., \( \text{Var}(X) = E[(X - E[X])^2] \).

Binomial Distribution: The Binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. It is defined by two parameters: \(n\) (the number of trials) and \(p\) (the probability of success on a single trial).

Mean of Binomial Distribution: The mean (expected value) of a Binomial distribution \(X \sim B(n, p)\) is \(E[X] = np\).

Expected Value of Y: Although not required for variance, we can also find the expected value of Y. Using the property \(E[aX + b] = aE[X] + b\):

\( E[Y] = E[100 - X] = 100 - E[X] \)

\( E[X] = np = 100 \cdot \frac{4}{5} = 80 \)

\( E[Y] = 100 - 80 = 20 \)

The fact that \(X\) follows a Binomial distribution means that \(Y = 100 - X\) actually follows a Binomial distribution as well, specifically \(Y \sim B(n=100, p'=1-p)\). Here, \(p' = 1 - \frac{4}{5} = \frac{1}{5}\). The variance of this distribution would be \( n \cdot p' \cdot (1-p') = 100 \cdot \frac{1}{5} \cdot (1 - \frac{1}{5}) = 100 \cdot \frac{1}{5} \cdot \frac{4}{5} = 16 \), which matches the result obtained using variance properties.