In a binomial distribution, if the mean is 6 and the standard deviation is \sqrt{2}, then what are the values of the parameters n and p respectively?

Explanation

For a binomial distribution, the mean is \mu = np = 6 and the variance is \sigma^2 = npq = (\sqrt{2})^2 = 2. Dividing the variance by the mean gives q = \frac{npq}{np} = \frac{2}{6} = \frac{1}{3}. Since p + q = 1, we have p = 1 - \frac{1}{3} = \frac{2}{3}. Substituting p back into the mean equation yields n(\frac{2}{3}) = 6, which solves to n = 9.

Related questions on Statistics & Probability

• Let x be the mean of squares of first n natural numbers and y be the square of mean of first n natural numbers. If $\frac{x}{y}=\fra...
• What is the probability of getting a composite number in the list of natural numbers from 1 to 50?
• Two numbers x and y are chosen at random from a set of first 10 natural numbers. What is the probability that (x+y) is divisible by 4?
• A number x is chosen at random from first n natural numbers. What is the probability that the number chosen satisfies $x+\frac{1}{x} \gt...
• Three fair dice are tossed once. What is the probability that they show different numbers that are in AP?