Understanding Binomial Distribution Mean and Variance

A binomial distribution is a discrete probability distribution that describes the probability of obtaining a certain number of successes in a fixed number of independent Bernoulli trials. Two key parameters define a binomial distribution: the number of trials ($n$) and the probability of success in a single trial ($p$).

For a binomial distribution with parameters $n$ and $p$, the mean ($\mu$) and variance ($\sigma^2$) are given by specific formulas:

• Mean ($\mu$) = $np$
• Variance ($\sigma^2$) = $np(1-p)$

In this problem, we are given the mean and variance of a binomial distribution and need to find the value of $n$, the number of trials.

Solving for Binomial Parameters

We are given the following information:

• Mean, $np = 5$
• Variance, $np(1-p) = 4$

We have a system of two equations with two unknowns, $n$ and $p$. We can use these equations to solve for $p$ first, and then use the value of $p$ to find $n$.

Let's use the given equations:

1. $\qquad np = 5$
2. $\qquad np(1-p) = 4$

Substitute the value of $np$ from equation (1) into equation (2):

$\qquad 5(1-p) = 4$

Now, we can solve for $p$:

$\qquad 1-p = \frac{4}{5}$

$\qquad p = 1 - \frac{4}{5}$

$\qquad p = \frac{5-4}{5}$

$\qquad p = \frac{1}{5}$

The probability of success, $p$, is $\frac{1}{5}$. Since $p$ represents a probability, its value must be between 0 and 1 ($0 \le p \le 1$). Our calculated value $p=1/5$ satisfies this condition.

Now that we have the value of $p$, we can substitute it back into equation (1) to find the value of $n$:

$\qquad np = 5$

$\qquad n \left(\frac{1}{5}\right) = 5$

To solve for $n$, multiply both sides by 5:

$\qquad n = 5 \times 5$

$\qquad n = 25$

The value of $n$, the number of trials, is 25. Since $n$ represents the number of trials, it must be a positive integer. Our calculated value $n=25$ is a positive integer.

Verifying the Binomial Parameters

With $n=25$ and $p=1/5$, let's check if the mean and variance match the given values:

• Mean = $np = 25 \times \frac{1}{5} = \frac{25}{5} = 5$. This matches the given mean.
• Variance = $np(1-p) = 25 \times \frac{1}{5} \times (1 - \frac{1}{5}) = 5 \times \frac{4}{5} = \frac{20}{5} = 4$. This matches the given variance.

The values $n=25$ and $p=1/5$ are consistent with the given mean and variance of the binomial distribution.

The value of $n$ is 25.

Revision Table: Binomial Distribution Formulas

Additional Information: Properties of Binomial Distribution

A binomial distribution arises from a sequence of Bernoulli trials. A Bernoulli trial is a single experiment with only two possible outcomes: success or failure. Key properties of a binomial distribution include:

• There are a fixed number of trials, denoted by $n$.
• Each trial is independent of the others.
• For each trial, there are only two possible outcomes: "success" and "failure".
• The probability of success, denoted by $p$, is constant for every trial. The probability of failure is $q = 1-p$.
• The random variable $X$ represents the number of successes in $n$ trials.

Understanding the mean and variance formulas is crucial for working with binomial distributions, as they allow us to characterize the distribution's central tendency and spread based on the parameters $n$ and $p$. In this problem, we used the inverse approach, starting from the mean and variance to find the parameters.