The mean and variance of a binomial variate X | 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 $np = 16/5$ and the variance is $npq = 48/25$.Dividing the variance by the mean, we get $q = (48/25) / (1

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(A) For a binomial distribution, the mean is $np = 16/5$ and the variance is $npq = 48/25$.Dividing the variance by the mean, we get $q = (48/25) / (16/5) = (48/25) 	imes (5/16) = 3/5$.Since $p + q = 1$, we have $p = 1 - 3/5 = 2/5$.Substituting $p$ into the mean equation: $n(2/5) = 16/5 \Rightarrow n = 8$.We need to find $P(X > 1) = 1 - P(X = 0) - P(X = 1)$.$P(X = 0) = ^8C_0 (2/5)^0 (3/5)^8 = (3/5)^8$.$P(X = 1) = ^8C_1 (2/5)^1 (3/5)^7 = 8 	imes (2/5) 	imes (3/5)^7 = (16/5) 	imes (3/5)^7$.Thus, $P(X > 1) = 1 - (3/5)(3/5)^7 - (16/5)(3/5)^7 = 1 - (3/5 + 16/5)(3/5)^7 = 1 - (19/5)(3/5)^7$.Comparing this with $1 - K(3/5)^7$, we get $K = 19/5$.Therefore, $5K = 5 	imes (19/5) = 19$.

The mean and variance of a binomial variate $X$ are $16/5$ and $48/25$ respectively. If $P(X > 1) = 1 - K (3/5)^7$, then $5 K =$

Similar Questions

In a game,a pair of dice is rolled $24$ times. If a person wins the game by not getting $6$ on both the dice in any one of the $24$ rolls,then the probability that a person wins the game is

In a binomial distribution $B(n, p = 1/4)$,if the probability of at least one success is $\geq 9/10$,then $n \geq$ ?

The probability that a student is not a swimmer is $1/5$. What is the probability that $4$ out of $5$ students are swimmers?

$A$ 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:

$A$ die is thrown $6$ times. If 'getting an odd number' is a success,what is the probability of at least $5$ successes?

Vedclass Test Series

Exam Paper Generator

Online Exam Module