The mean and variance of a binomial distribut | 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 $np = 4$ and the variance is given by $npq = 2$. Dividing the variance by the mean,we get $q = \f

Vedclass · Accepted Answer

$\frac{7}{64}$

Vedclass · Answer

(A) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$. 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 exactly $x$ successes is given by $P(X=x) = {}^{n}C_{x} p^{x} q^{n-x}$. For $x = 2$,$P(X=2) = {}^{8}C_{2} \left(\frac{1}{2}ight)^{2} \left(\frac{1}{2}ight)^{8-2} = {}^{8}C_{2} \left(\frac{1}{2}ight)^{8}$. Calculating the value,${}^{8}C_{2} = \frac{8 	imes 7}{2 	imes 1} = 28$. Thus,$P(X=2) = 28 	imes \frac{1}{256} = \frac{28}{256} = \frac{7}{64}$.

The mean and variance of a binomial distribution are $4$ and $2$ respectively. Then the probability of exactly two successes is

Similar Questions

$A$ student is given $6$ questions in an examination with true or false type of answers. If he writes $4$ or more correct answers,he passes in the examination. The probability that he passes in the examination is

For Binomial distribution $B \left(10, \frac{1}{2} \right)$,if $P(X \leq 2) = m \left( \frac{1}{2} \right)^{10}$,then $m =$ . . . . . . .

$A$ die is tossed thrice. $A$ success is defined as getting $1$ or $6$ on a toss. Find the mean and the variance of the number of successes.

Suppose that $90 \%$ of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module