Let X have a binomial distribution B(n, p) su | vedclass

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

Question

(B) Let $\mu = np$ be the mean and $\sigma^2 = npq$ be the variance of the binomial distribution $X \sim B(n, p)$.Given $\mu + \sigma^2 = 24$ and $\mu

Vedclass · Accepted Answer

$529$

Vedclass · Answer

(B) Let $\mu = np$ be the mean and $\sigma^2 = npq$ be the variance of the binomial distribution $X \sim B(n, p)$.Given $\mu + \sigma^2 = 24$ and $\mu \sigma^2 = 128$.Let $x = \mu$ and $y = \sigma^2$. Then $x + y = 24$ and $xy = 128$.The quadratic equation $t^2 - 24t + 128 = 0$ has roots $t = 16$ and $t = 8$.Since $\mu > \sigma^2$ (as $q Thus,$np = 16$ and $npq = 8$. Dividing these gives $q = \frac{8}{16} = \frac{1}{2}$.Since $p = 1 - q$,we have $p = \frac{1}{2}$.Then $n 	imes \frac{1}{2} = 16$,so $n = 32$.We need to find $P(X > n - 3) = P(X > 29) = P(X = 30) + P(X = 31) + P(X = 32)$.$P(X = r) = {}^{n}C_r p^r q^{n-r} = {}^{32}C_r (\frac{1}{2})^r (\frac{1}{2})^{32-r} = \frac{{}^{32}C_r}{2^{32}}$.So,$P(X > 29) = \frac{{}^{32}C_{30} + {}^{32}C_{31} + {}^{32}C_{32}}{2^{32}} = \frac{k}{2^{32}}$.Thus,$k = {}^{32}C_{30} + {}^{32}C_{31} + {}^{32}C_{32} = {}^{32}C_2 + {}^{32}C_1 + {}^{32}C_0$.$k = \frac{32 	imes 31}{2} + 32 + 1 = 496 + 32 + 1 = 529$.

Let $X$ have a binomial distribution $B(n, p)$ such that the sum and the product of the mean and variance of $X$ are $24$ and $128$ respectively. If $P(X > n - 3) = \frac{k}{2^n}$,then $k$ is equal to.

Similar Questions

The probability of a man hitting a target is $\frac{2}{5}$. He fires at the target $k$ times ($k$ is a given number). The minimum value of $k$ such that the probability of hitting the target at least once is more than $\frac{7}{10}$ is:

The mean and standard deviation of a binomial variate $X$ are $4$ and $\sqrt{3}$ respectively. Then $P(X \geq 1)$ is equal to

$A$ pair of fair dice is thrown $4$ times. If getting the same number on both dice is considered as a success,then the probability of exactly two successes is:

$A$ contest consists of predicting the results (win,draw,or defeat) of $7$ football matches. $A$ sent his entry by predicting at random. The probability that his entry will contain exactly $4$ correct predictions 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