Let X sim B(n, p) with mean mu and variance s | vedclass

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

Question

(D) For a binomial distribution $X \sim B(n, p)$,the mean $\mu = np$ and variance $\sigma^2 = npq$,where $q = 1-p$.Given $\mu = 2\sigma^2$,we have $np

Vedclass · Accepted Answer

$\frac{15}{16}$

Vedclass · Answer

(D) For a binomial distribution $X \sim B(n, p)$,the mean $\mu = np$ and variance $\sigma^2 = npq$,where $q = 1-p$.Given $\mu = 2\sigma^2$,we have $np = 2npq$,which implies $1 = 2q$,so $q = \frac{1}{2}$ and $p = 1 - q = \frac{1}{2}$.Given $\mu + \sigma^2 = 3$,we substitute $\mu = 2\sigma^2$ to get $3\sigma^2 = 3$,so $\sigma^2 = 1$.Since $\sigma^2 = npq = n(\frac{1}{2})(\frac{1}{2}) = \frac{n}{4} = 1$,we find $n = 4$.Thus,$X \sim B(4, \frac{1}{2})$.We need to calculate $P(X \leq 3) = 1 - P(X = 4)$.$P(X = 4) = \binom{4}{4} p^4 q^0 = 1 	imes (\frac{1}{2})^4 	imes 1 = \frac{1}{16}$.Therefore,$P(X \leq 3) = 1 - \frac{1}{16} = \frac{15}{16}$.

Let $X \sim B(n, p)$ with mean $\mu$ and variance $\sigma^2$. If $\mu=2 \sigma^2$ and $\mu+\sigma^2=3$,then $P(X \leq 3)=$

Similar Questions

The sum and product of the mean and variance of a binomial distribution are $82.5$ and $1350$ respectively. The number of trials in the binomial distribution is:

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

Two dice are thrown successively $4$ times and getting a doublet is called a success. The probability of getting at least $1$ success is (in $/1296$)

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

If $8$ coins are tossed simultaneously,then the probability of getting at least $6$ heads is

Vedclass Test Series

Exam Paper Generator

Online Exam Module