Binomial Distribution: Finding Parameters n and p
This problem requires us to find the parameters of a binomial distribution, specifically the number of trials (\(n\)) and the probability of success (\(p\)), given its mean (\(\mu\)) and standard deviation (\(\sigma\)).
Understanding Binomial Distribution Formulas
For a binomial distribution, the following formulas relate the parameters (\(n\), \(p\)) to the mean (\(\mu\)) and standard deviation (\(\sigma\)):
- Mean: \(\mu = np\)
- Variance: \(\sigma^2 = np(1-p)\)
- Standard Deviation: \(\sigma = \sqrt{np(1-p)}\)
Applying Given Values
We are given:
- Mean, \(\mu = 6\)
- Standard Deviation, \(\sigma = \sqrt{2}\)
From the standard deviation, we can calculate the variance:
Variance, \(\sigma^2 = (\sqrt{2})^2 = 2\)
Step-by-Step Calculation
Using the formulas and the given values, we can set up a system of equations:
- From the mean formula: \(np = 6\)
- From the variance formula: \(np(1-p) = 2\)
Now, we can solve these equations:
Step 1: Substitute the mean into the variance equation.
We know \(np = 6\). Substitute this into the second equation:
\(6(1-p) = 2\)Step 2: Solve for \(p\).
Divide both sides by 6:
\(1-p = \frac{2}{6}\) \(1-p = \frac{1}{3}\)Rearrange to find \(p\):
\(p = 1 - \frac{1}{3}\) \(p = \frac{3}{3} - \frac{1}{3}\) \(p = \frac{2}{3}\)Step 3: Solve for \(n\).
Substitute the value of \(p = \frac{2}{3}\) back into the mean equation (\(np = 6\)):
\(n \left( \frac{2}{3} \right) = 6\)Multiply both sides by \(\frac{3}{2}\) to solve for \(n\):
\(n = 6 \times \frac{3}{2}\) \(n = \frac{18}{2}\) \(n = 9\)Conclusion
The calculated values for the parameters are \(n=9\) and \(p=\frac{2}{3}\).
Therefore, the values of the parameters \(n\) and \(p\) respectively are 9 and 2/3.
Similar Questions
- The occurrence of a disease in an industry is such that the workers have 20% chance of suffering from it. What is the probability that out of 6 workers chosen at random, 4 or more will suffer from the disease?
- For a Binomial distribution with mean 6 and standard deviation \(\sqrt{2}\), what is the value of P(X = 0) ?
Let X be a random variable following a binomial distribution whose mean and variance are 200 and 160 respectively. What is the value of the number of trials n?
Important Questions from Binomial Distribution
In a Binomial distribution, the mean is three times its variance. What is the probability of exactly 3 successes out of 5 trials?
If mean and variance of a Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is
For Binomial distribution, n = 10 and p = 0.6, E(X 2) (second moment about origin) is:
Find out the fallacy if any in the statement:
“The mean and the variance of a binomial distribution is 16.2 and 29.4 respectively.”
If log e(x) is normally distributed with mean 1 and variance 4, then P(0.5 < x < 2) is:
(where the area between z = 0 and z = 0.25 is 0.0987 and the area between z = 0 and z = 0.5 is 0.1915)
