If the mean and the variance of a Binomial va | vedclass

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

Question

(B) Given that the mean $np = 2$ and the variance $npq = 1$. Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{1}{2}$,which implies $q

Vedclass · Accepted Answer

$\frac{11}{16}$

Vedclass · Answer

(B) Given that the mean $np = 2$ and the variance $npq = 1$. Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{1}{2}$,which implies $q = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n \left( \frac{1}{2} ight) = 2$,so $n = 4$. We need to find $P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)$. Alternatively,$P(X > 1) = 1 - [P(X = 0) + P(X = 1)]$. $P(X = 0) = {}^4 C_0 \left( \frac{1}{2} ight)^0 \left( \frac{1}{2} ight)^4 = 1 	imes 1 	imes \frac{1}{16} = \frac{1}{16}$. $P(X = 1) = {}^4 C_1 \left( \frac{1}{2} ight)^1 \left( \frac{1}{2} ight)^3 = 4 	imes \frac{1}{2} 	imes \frac{1}{8} = \frac{4}{16}$. Therefore,$P(X > 1) = 1 - \left( \frac{1}{16} + \frac{4}{16} ight) = 1 - \frac{5}{16} = \frac{11}{16}$.

If the mean and the variance of a Binomial variate $X$ are $2$ and $1$ respectively,then the probability that $X$ takes a value greater than $1$ is equal to

Similar Questions

Let in a Binomial distribution,consisting of $5$ independent trials,probabilities of exactly $1$ and $2$ successes be $0.4096$ and $0.2048$ respectively. Then the probability of getting exactly $3$ successes is equal to ....... .

If $7$ different balls are distributed among $4$ different boxes,then the probability that the first box contains $3$ balls is

If the mean and the variance of a Binomial variate $X$ are $2$ and $1$ respectively,then the probability that $X$ takes a value greater than or equal to $1$ is

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. The mean of the number of kings is:

If $X$ is a binomial variate with mean $6$ and variance $2$,then the value of $P(5 \leq X \leq 7)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module