Binomial Distribution Properties Recap

This problem involves a random variable (\(x\)) following a binomial distribution. Let's recall the key properties of a binomial distribution:

• It has two parameters: \(n\) (the number of trials) and \(p\) (the probability of success in a single trial).
• The mean (\(\mu\)) is given by \(\mu = np\).
• The variance (\(\sigma^2\)) is given by \(\sigma^2 = np(1-p)\).
• The probability mass function (PMF) is \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(k\) is the number of successes.

Binomial Distribution Parameters Calculation

We are given the following information:

• Mean: \(\mu = 5\)
• Variance: \(\sigma^2 = 4\)

Using the formulas for mean and variance, we can set up a system of equations to find the parameters \(n\) and \(p\):

1. From the mean: \(np = 5\)
2. From the variance: \(np(1-p) = 4\)

Now, substitute the value of \(np\) from equation (1) into equation (2):

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

Solve for \(p\):

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

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

Next, substitute the value of \(p\) back into equation (1) to find \(n\):

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

\(n = 5 \times 5 = 25\)

So, the parameters of the binomial distribution are \(n=25\) and \(p=\frac{1}{5}\).

Probability Calculation for P(X=3)

We need to find the probability \(P(X=3)\) using the binomial PMF with \(n=25\), \(p=\frac{1}{5}\), and \(k=3\). First, let's find \(1-p\):

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

Now calculate the binomial coefficient \(\binom{n}{k} = \binom{25}{3}\):

\(\binom{25}{3} = \frac{25!}{3!(25-3)!} = \frac{25!}{3!22!} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 100 \times 23 = 2300\)

Now, apply the PMF formula:

\(P(X=3) = \binom{25}{3} p^3 (1-p)^{25-3}\)

\(P(X=3) = 2300 \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^{22}\)

\(P(X=3) = 2300 \times \frac{1^3}{5^3} \times \frac{4^{22}}{5^{22}} = 2300 \times \frac{1}{125} \times \frac{4^{22}}{5^{22}}\)

\(P(X=3) = 2300 \times \frac{4^{22}}{5^3 \times 5^{22}} = 2300 \times \frac{4^{22}}{5^{25}}\)

Solving for Lambda (\(\lambda\))

We are given the equation:

\(5^{23} P(X = 3) = \lambda 4^{\lambda}\)

Substitute the expression for \(P(X=3)\) we just calculated:

\(5^{23} \left( 2300 \times \frac{4^{22}}{5^{25}} \right) = \lambda 4^{\lambda}\)

Simplify the left side:

\(\frac{5^{23}}{5^{25}} \times 2300 \times 4^{22} = \lambda 4^{\lambda}\)

\(\frac{1}{5^2} \times 2300 \times 4^{22} = \lambda 4^{\lambda}\)

\(\frac{2300}{25} \times 4^{22} = \lambda 4^{\lambda}\)

Calculate \(\frac{2300}{25}\):

\(\frac{2300}{25} = 92\)

So the equation becomes:

\(92 \times 4^{22} = \lambda 4^{\lambda}\)

We can rewrite \(92\) as \(23 \times 4\):

\((23 \times 4) \times 4^{22} = \lambda 4^{\lambda}\)

Using the exponent rule \(a^m \times a^n = a^{m+n}\):

\(23 \times 4^{1+22} = \lambda 4^{\lambda}\)

\(23 \times 4^{23} = \lambda 4^{\lambda}\)

By comparing the structure of both sides of the equation, we can see that \(\lambda\) must be 23.

Therefore, \(\lambda = 23\).