Question

If the mean and variance of a binomial variate are 12 and 4, then the distribution is \(\ldots\)

Short answer

The distribution is a binomial distribution with \( n = 18 \) and \( p = \frac{2}{3} \).

Step-by-step solution

Understanding the Binomial Distribution

The binomial distribution is characterized by two parameters: the number of trials \( n \) and the probability of success \( p \). The mean of a binomial distribution is given by \( \mu = n \cdot p \) and the variance by \( \sigma^2 = n \cdot p \cdot (1-p) \). We will use these formulas to find \( n \) and \( p \).

Setting up the Equations

From the problem statement, the mean \( \mu = 12 \) and the variance \( \sigma^2 = 4 \). Thus, we have the two equations: \( n \cdot p = 12 \) and \( n \cdot p \cdot (1-p) = 4 \).

Solving for n in Terms of p

From Equation 1, we can express \( n \) in terms of \( p \): \( n = \frac{12}{p} \).

Substituting and Solving for p

Substituting \( n = \frac{12}{p} \) into the variance equation, we get \[ \frac{12}{p} \cdot p \cdot (1-p) = 4. \] Simplifying gives \( 12(1-p) = 4 \). Solve for \( p \) by rearranging: \( 12 - 12p = 4 \), which simplifies to \( 12p = 8 \), thus \( p = \frac{2}{3} \).

Finding the Value of n

Now that we know \( p = \frac{2}{3} \), substitute back into \( n = \frac{12}{p} \): \( n = \frac{12}{\frac{2}{3}} = 18 \).

Conclusion

The binomial distribution is characterized by \( n = 18 \) and \( p = \frac{2}{3} \). Thus, the parameters of the distribution are \( n = 18 \) and \( p = \frac{2}{3} \).

Key concepts

Mean and Variance in Binomial Distribution

Binomial distribution is an important probability distribution in the field of statistics and probability theory. It primarily describes the number of successes in a fixed number of independent Bernoulli trials, which are experiments with two possible outcomes: success or failure. To understand it better, let's focus on two key concepts: mean and variance.

The mean of a binomial distribution, denoted by \( \mu \), is calculated using the formula \( \mu = n \cdot p \), where \( n \) is the number of trials and \( p \) is the probability of success in one trial. This formula essentially gives you an expectation of the number of successes.

Variance, another vital measurement, helps us understand the variability or spread in the number of successes. Calculated by the formula \( \sigma^2 = n \cdot p \cdot (1-p) \), variance considers both the probability of success \( p \) and the probability of failure \( 1-p \). This calculation tells us how much the actual number of successes might differ from the mean.

Solving Binomial Parameters

To fully define a binomial distribution, we need two parameters: the number of trials \( n \) and the probability of success \( p \). When given both the mean \( \mu \) and variance \( \sigma^2 \), we can find these parameters through a series of calculations.

Start with the two equations that relate the mean and variance to the parameters:

• Mean: \( n \cdot p = 12 \)
• Variance: \( n \cdot p \cdot (1-p) = 4 \)

Through substitution, you can express \( n \) in terms of \( p \), leading to a simplified equation that helps us solve for \( p \). For example, substituting \( n = \frac{12}{p} \) in the variance equation provides an equation in terms of \( p \) only. After solving, you find out that \( p = \frac{2}{3} \).

Next, using this value of \( p \), substitute back to find \( n \): \( n = \frac{12}{\frac{2}{3}} = 18 \). Therefore, the binomial distribution has these parameters: 18 trials with a probability of success of \( \frac{2}{3} \).

Probability of Success in Binomial Distribution

In any binomial distribution, the probability of success \( p \) is a fundamental aspect that dictates the likelihood of a successful outcome in a single trial. Solving for \( p \) means determining this likelihood when other parameters or statistics of the distribution are given.

From the formula involving mean: \( n \cdot p = 12 \), and the variance: \( n \cdot p \cdot (1-p) = 4 \), one can deduce \( p \) by eliminating one variable and solving for the other. This conversion and solving process shows how mathematical relationships between these numbers express probabilities.

Here, if you solve the equation \( 12 - 12p = 4 \) after substitution into variance, you'll find \( p = \frac{2}{3} \). This implies that in every trial, the chance of success is approximately 66.67%. Understanding the probability of success is crucial, as it affects the shape and behavior of the entire distribution.