In a Binomial distribution B(n, p),the sum an | vedclass

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

Question

(C) For a Binomial distribution $B(n, p)$,Mean $\mu = np$ and Variance $\sigma^2 = npq$,where $q = 1-p$. Given: Sum: $np + npq = 5 \Rightarrow np(1+q)

Vedclass · Accepted Answer

$52$

Vedclass · Answer

(C) For a Binomial distribution $B(n, p)$,Mean $\mu = np$ and Variance $\sigma^2 = npq$,where $q = 1-p$. Given: Sum: $np + npq = 5 \Rightarrow np(1+q) = 5$ ...$(i)$ Product: $(np)(npq) = n^2p^2q = 6$ ...(ii) From $(i)$,$np = \frac{5}{1+q}$. Substituting this into (ii): $\left(\frac{5}{1+q}\right)^2 q = 6 \Rightarrow 25q = 6(1+q)^2 \Rightarrow 25q = 6(1+2q+q^2) \Rightarrow 6q^2 - 13q + 6 = 0$. Solving the quadratic equation: $6q^2 - 9q - 4q + 6 = 0 \Rightarrow 3q(2q-3) - 2(2q-3) = 0 \Rightarrow (3q-2)(2q-3) = 0$. Since $q Substituting $q = \frac{2}{3}$ into $(i)$: $np(1 + \frac{2}{3}) = 5 \Rightarrow np(\frac{5}{3}) = 5 \Rightarrow np = 3$. Since $p = \frac{1}{3}$,$n(\frac{1}{3}) = 3 \Rightarrow n = 9$. Finally,$6(n+p-q) = 6(9 + \frac{1}{3} - \frac{2}{3}) = 6(9 - \frac{1}{3}) = 54 - 2 = 52$.

In a Binomial distribution $B(n, p)$,the sum and product of the mean and the variance are $5$ and $6$ respectively,then $6(n+p-q)=$

Similar Questions

Let $\lim _{t \rightarrow 0}(1+5 t)^{\frac{1}{t}}=K$ and $X$ be the random variable representing the number of successes in $100$ independent trials. If the probability of success in each trial is $0.05$, then the probability of getting at least one success is

$A$ multiple-choice examination consists of $5$ questions. Each question has $3$ alternative answers,of which only $1$ is correct. What is the probability that a student will get $4$ or more correct answers?

Five cards are drawn successively with replacement from a well-shuffled deck of $52$ cards. What is the probability that none is a spade?

In a binomial distribution,the mean is $15$ and the variance is $10$. Then the parameter $n$ is:

In a random experiment,two dice are thrown and the sum of the numbers appeared on them is recorded. This experiment is repeated $9$ times. If the probability that a sum of $6$ appears at least once is $P_1$ and a sum of $8$ appears at least once is $P_2$,then $P_1 : P_2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module