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 \(p\):
- From the mean: \(np = 5\)
- 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, \(\lambda = 23\).
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Code:
If two regression coefficients are 0.8 and 1.2, which one of the following is the value of coefficient of correlation?
