Understanding Binomial Distribution Mean and Variance

A binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success.

For a binomial distribution \(B(x, n, p)\), where:

• \(n\) is the number of trials
• \(p\) is the probability of success in a single trial
• \(x\) is the number of successes

The mean (\(\mu\)) and variance (\(\sigma^2\)) are given by the formulas:

• Mean (\(\mu\)) = \(np\)
• Variance (\(\sigma^2\)) = \(npq\), where \(q = 1 - p\) is the probability of failure.

Calculating Parameters from Mean and Variance

We are given the mean and variance of a binomial distribution:

• Mean = 4
• Variance = \(\dfrac{4}{3}\)

Using the formulas, we can set up equations:

1. \(np = 4\)
2. \(npq = \dfrac{4}{3}\)

Substitute the first equation into the second equation:

\(4q = \dfrac{4}{3}\)

Now, solve for \(q\):

\(q = \dfrac{4}{3} \times \dfrac{1}{4}\)

\(q = \dfrac{1}{3}\)

Since \(p + q = 1\), we can find \(p\):

\(p = 1 - q\)

\(p = 1 - \dfrac{1}{3}\)

\(p = \dfrac{3 - 1}{3} = \dfrac{2}{3}\)

Now that we have \(p\), we can find \(n\) using the equation \(np = 4\):

\(n \times \dfrac{2}{3} = 4\)

\(n = 4 \times \dfrac{3}{2}\)

\(n = \dfrac{12}{2} = 6\)

So, the binomial distribution is \(B(x, 6, \dfrac{2}{3})\), with \(n=6\) and \(p=\dfrac{2}{3}\).

Finding the Probability of 2 Successes

The probability of getting exactly \(k\) successes in \(n\) trials for a binomial distribution is given by the probability mass function (PMF):

\(P(x=k) = \binom{n}{k} p^k q^{n-k}\)

Where \(\binom{n}{k} = \dfrac{n!}{k!(n-k)!}\) is the binomial coefficient.

We need to find the probability of getting 2 successes, so \(k=2\). We have \(n=6\), \(p=\dfrac{2}{3}\), and \(q=\dfrac{1}{3}\).

Substitute these values into the PMF formula:

\(P(x=2) = \binom{6}{2} \left(\dfrac{2}{3}\right)^2 \left(\dfrac{1}{3}\right)^{6-2}\)

\(P(x=2) = \binom{6}{2} \left(\dfrac{2}{3}\right)^2 \left(\dfrac{1}{3}\right)^4\)

First, calculate the binomial coefficient \(\binom{6}{2}\):

\(\binom{6}{2} = \dfrac{6!}{2!(6-2)!} = \dfrac{6!}{2!4!} = \dfrac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(4 \times 3 \times 2 \times 1)} = \dfrac{6 \times 5}{2 \times 1} = \dfrac{30}{2} = 15\)

Next, calculate the power terms:

\(\left(\dfrac{2}{3}\right)^2 = \dfrac{2^2}{3^2} = \dfrac{4}{9}\)

\(\left(\dfrac{1}{3}\right)^4 = \dfrac{1^4}{3^4} = \dfrac{1}{81}\)

Now, multiply these values together to find \(P(x=2)\):

\(P(x=2) = 15 \times \dfrac{4}{9} \times \dfrac{1}{81}\)

\(P(x=2) = \dfrac{15 \times 4 \times 1}{9 \times 81}\)

\(P(x=2) = \dfrac{60}{729}\)

To simplify the fraction, we can divide the numerator and denominator by their greatest common divisor. Both 60 and 729 are divisible by 3.

\(60 \div 3 = 20\)

\(729 \div 3 = 243\)

So, \(P(x=2) = \dfrac{20}{243}\).

This is the probability of getting exactly 2 successes in this binomial distribution.

Summary of Results

Based on the given mean and variance, we found the parameters of the binomial distribution and calculated the required probability.

Revision Table: Binomial Distribution Concepts

Additional Information: Properties of Binomial Distribution

The binomial distribution is a fundamental concept in probability and statistics. Understanding its properties is crucial for various applications.

• The shape of the binomial distribution depends on the values of \(n\) and \(p\). It can be symmetric (when \(p=0.5\)) or skewed.
• As \(n\) becomes large, the binomial distribution can be approximated by other distributions like the Poisson distribution (if \(p\) is small) or the normal distribution (if \(np > 5\) and \(n(1-p) > 5\)).
• The sum of probabilities for all possible numbers of successes (from 0 to n) must equal 1.
• The binomial distribution is widely used in quality control, medical research, and social sciences.