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

(B) For a binomial distribution,the mean is given by $\mu = np$ and the variance is given by $\sigma^2 = npq$,where $p + q = 1$.Given $\mu = 4$ and $\

Vedclass · Accepted Answer

${}^{16}C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}$

Vedclass · Answer

(B) For a binomial distribution,the mean is given by $\mu = np$ and the variance is given by $\sigma^2 = npq$,where $p + q = 1$.Given $\mu = 4$ and $\sigma^2 = 3$,we have:$np = 4$ and $npq = 3$.Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{3}{4}$.Since $p = 1 - q$,we have $p = 1 - \frac{3}{4} = \frac{1}{4}$.Substituting $p = \frac{1}{4}$ into $np = 4$,we get $n \left( \frac{1}{4} \right) = 4$,which implies $n = 16$.The probability of getting exactly $x$ successes is given by $P(X = x) = {}^nC_x p^x q^{n-x}$.For $x = 6$,$P(X = 6) = {}^{16}C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{16-6} = {}^{16}C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}$.Thus,the correct option is $B$.

The mean and variance of a binomial distribution are $4$ and $3$ respectively. Then,the probability of getting exactly six successes in this distribution is:

Similar Questions

The probability that an event will fail to happen is $0.05$. The probability that the event will take place on $4$ consecutive occasions is

One hundred identical coins,each with probability $p$ of showing up heads,are tossed once. If $0 < p < 1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins,then the value of $p$ is

If $X \sim B(n, p)$ with $n = 10, p = 0.4$,then $E(X^2) =$

For an initial screening of an entrance exam,a candidate is given $50$ problems to solve. If the probability that the candidate can solve any problem is $\frac{4}{5}$,then the probability that he is unable to solve less than $2$ problems is:

$A$ binomial random variable $X$ satisfies $9 \cdot P(X=4) = P(X=2)$ when $n=6$. Then $p$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module