If the mean and variance of a binomial distri | 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 $\mu = np = 4$ and the variance is given by $\sigma^2 = npq = 2$. Dividing the variance by the me

Vedclass · Accepted Answer

$\frac{28}{256}$

Vedclass · Answer

(A) For a binomial distribution,the mean is given by $\mu = np = 4$ and the variance is given by $\sigma^2 = npq = 2$. Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{2}{4} = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n 	imes \frac{1}{2} = 4$,which implies $n = 8$. The probability of getting $x$ successes is given by $P(X=x) = { }^n C_x p^x q^{n-x}$. For $x = 2$,$P(X=2) = { }^8 C_2 \left(\frac{1}{2}ight)^2 \left(\frac{1}{2}ight)^6 = \frac{8 	imes 7}{2 	imes 1} 	imes \left(\frac{1}{2}ight)^8$. $P(X=2) = 28 	imes \frac{1}{256} = \frac{28}{256}$.

If the mean and variance of a binomial distribution are $4$ and $2$ respectively,then the probability of getting $2$ successes is:

Similar Questions

If a fair coin is tossed $10$ times,find the probability of exactly six heads.

Let a random variable $X$ have a Binomial distribution with mean $8$ and variance $4$. If $P(X \leqslant 2) = \frac{k}{2^{16}}$,then $k$ is equal to:

One out of $9$ ships is likely to sink when they are set on sail. When $6$ ships are set on sail,the probability that exactly $3$ of them will not arrive safely is

Suppose $X$ follows a binomial distribution with parameters $n$ and $p$,where $0 < p < 1$. If $\frac{P(X=r)}{P(X=n-r)}$ is independent of $n$ for every $r$,then $p$ is equal to

The probability that a man can hit a target is $\frac{3}{4}$. He tries $5$ times. The probability that he will hit the target at least three times is

Vedclass Test Series

Exam Paper Generator

Online Exam Module