7 coins are tossed simultaneously and the num | vedclass

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

Question

(C) For a binomial distribution with $n=7$ and $p=q=\frac{1}{2}$:$\mu = np = 7 	imes \frac{1}{2} = \frac{7}{2}$$\sigma^2 = npq = 7 	imes \frac{1}{2}

Vedclass · Accepted Answer

$\frac{112}{5}$

Vedclass · Answer

(C) For a binomial distribution with $n=7$ and $p=q=\frac{1}{2}$:$\mu = np = 7 	imes \frac{1}{2} = \frac{7}{2}$$\sigma^2 = npq = 7 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{7}{4}$$P(X=3) = {}^{7}C_{3} 	imes (\frac{1}{2})^{3} 	imes (\frac{1}{2})^{4} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} 	imes (\frac{1}{2})^{7} = 35 	imes \frac{1}{128} = \frac{35}{128}$$\frac{\mu \sigma^2}{P(X=3)} = \frac{(\frac{7}{2}) 	imes (\frac{7}{4})}{\frac{35}{128}} = \frac{49}{8} 	imes \frac{128}{35} = \frac{7}{1} 	imes \frac{16}{5} = \frac{112}{5}$

$7$ coins are tossed simultaneously and the number of heads turned up is denoted by the random variable $X$. If $\mu$ is the mean and $\sigma^2$ is the variance of $X$,then $\frac{\mu \sigma^2}{P(X=3)}=$

Similar Questions

If $X \sim B(35, p)$ such that $7 P(X=0)=P(X=1)$,then the value of $\frac{P(X=15)}{P(X=20)}$ is equal to

In a binomial distribution,the difference between the mean and standard deviation is $3$ and the difference between their squares is $21$,then $P(x=1) : P(x=2) =$

From a lot containing $10$ defective and $90$ non-defective bulbs,$8$ bulbs are selected one by one with replacement. Then the probability of getting at least $7$ defective bulbs is:

In a random experiment of throwing $5$ coins,the number of heads is defined as a random variable. The mean of the random variable is

$A$ die is thrown $10$ times. The probability that an odd number will come up at least once is

Vedclass Test Series

Exam Paper Generator

Online Exam Module