Given X sim B(n, p),if E(X) = 4 and operatorn | 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 is given by $E(X) = np$ and the variance is given by $\operatorname{Var}(X) = npq$,where $q

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np$ and the variance is given by $\operatorname{Var}(X) = npq$,where $q = 1 - p$.Given $E(X) = np = 4$.Given $\operatorname{Var}(X) = npq = 2.4$.Substituting $np = 4$ into the variance equation: $4q = 2.4$.Solving for $q$: $q = \frac{2.4}{4} = 0.6 = \frac{3}{5}$.Since $p = 1 - q$,we have $p = 1 - 0.6 = 0.4 = \frac{2}{5}$.Now,substitute $p$ back into the mean equation: $n 	imes \frac{2}{5} = 4$.$n = 4 	imes \frac{5}{2} = 10$.Thus,the value of $n$ is $10$.

Given $X \sim B(n, p)$,if $E(X) = 4$ and $\operatorname{Var}(X) = 2.4$,then $n = $

Similar Questions

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. Let $X$ denote the random variable of the number of jacks obtained in the two drawn cards. Then $P(X=1) + P(X=2)$ equals

$A$ die is thrown thrice. If getting $1$ or $6$ in a single throw is considered as success, then the variance of the number of successes is

If a random variable $X \sim B\left(n=5, p=\frac{1}{3}\right)$,then $P(2 < X < 4) = $

$A$ fair coin is tossed $n$ times such that the probability of getting at least one head is at least $0.9$. Then the minimum value of $n$ is:

$A$ multiple choice examination has $5$ questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get $4$ or more correct answers just by guessing is

Vedclass Test Series

Exam Paper Generator

Online Exam Module