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:
Indicate the correct answer for the combination from the following regarding the conditions for the applicability of a binominal distribution:
(a) There are n independent trials
(b) Each trial has only two possible outcomes
(c) The probabilities of two outcomes do not remain constant
(d) The trials are independent
Which of the following options is correct?
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.”
