Let X and Y be two random variables such that X + Y = 100. If X follows Binomial distribution with parameters n = 100 and p = \(\frac{4}{5}\), what is the variance of Y?
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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
| Concept | Formula/Property | Application in Problem |
|---|---|---|
| Relationship X and Y | \(Y = 100 - X\) | Used to express Y in terms of X |
| Variance Property | \( \text{Var}(aX + b) = a^2 \text{Var}(X) \) | \( \text{Var}(100 - X) = (-1)^2 \text{Var}(X) \) |
| Binomial Variance | \( \text{Var}(X) = np(1-p) \) | Calculated Var(X) using \(n=100, p=4/5\) |
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.
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Important Questions from Binomial Distribution
Indicate the correct answer for the combination from the following regarding the conditions for the applicability of a binominal distribution:
(a) There are n independent trials
(b) Each trial has only two possible outcomes
(c) The probabilities of two outcomes do not remain constant
(d) The trials are independent
Which of the following options is correct?
In which of the following practical situations, Poisson Distribution can be used?
A. Number of customers arriving at the super markets per hour.
B. Number of typographical errors per page in a typed material.
C. Number of accidents taking place per day on a busy road.
D. Dice throwing problems.
E. Number of defective material say, blades, etc. in a packing manufactured by a good concern.
Choose the most appropriate answer from the options given below:
For Binomial distribution, n = 10 and p = 0.6, E(X 2) (second moment about origin) is:
The mean and variance of binomial distribution B (x, n, p) are 4 and \(\dfrac{4}{3}\) respectively. What is the probability of getting 2 successes?
Find out the fallacy if any in the statement:
“The mean and the variance of a binomial distribution is 16.2 and 29.4 respectively.”
