In a binomial probability distribution,the me | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a binomial distribution,the mean is given by $\mu = np = 3$ and the standard deviation is $\sigma = \sqrt{npq} = \frac{3}{2}$.Squaring the sta

Vedclass · Accepted Answer

$(\frac{3}{4} + \frac{1}{4})^{12}$

Vedclass · Answer

(A) For a binomial distribution,the mean is given by $\mu = np = 3$ and the standard deviation is $\sigma = \sqrt{npq} = \frac{3}{2}$.Squaring the standard deviation,we get $npq = \frac{9}{4}$.Dividing $npq$ by $np$,we find $q = \frac{npq}{np} = \frac{9/4}{3} = \frac{3}{4}$.Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{3}{4} = \frac{1}{4}$.Now,substituting $p = \frac{1}{4}$ into $np = 3$,we get $n(\frac{1}{4}) = 3$,which implies $n = 12$.The binomial distribution is given by $(q + p)^n = (\frac{3}{4} + \frac{1}{4})^{12}$.

In a binomial probability distribution,the mean is $3$ and the standard deviation is $\frac{3}{2}$. Then the probability distribution is:

Similar Questions

Let $X$ be the number of successes in $n$ independent Bernoulli trials with probability of success $p = \frac{3}{4}$. The least value of $n$ so that $P(X \ge 1) \ge 0.9375$ is . . . . . .

$A$ fair die is thrown $4$ times. If getting a prime number on the die is considered as a success, then the probability of getting no success at all is ...

If the mean and variance of a binomial distribution are $4$ and $\frac{4}{3}$ respectively, then $P(X=2)=$

$A$ test consists of $6$ multiple choice questions,each having $4$ alternative answers of which only one is correct. The number of ways in which a candidate answers all $6$ questions such that exactly $4$ of the answers are correct is:

When a coin is tossed $6$ times,the probability of getting more heads than tails is

Vedclass Test Series

Exam Paper Generator

Online Exam Module