The maximum value of the variance of a Binomi | vedclass

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

Question

(B) The variance of a Binomial distribution is given by $\sigma^2 = n p q$,where $q = 1 - p$. To find the maximum value of the variance,we express it

Vedclass · Accepted Answer

$\frac{n}{4}$

Vedclass · Answer

(B) The variance of a Binomial distribution is given by $\sigma^2 = n p q$,where $q = 1 - p$. To find the maximum value of the variance,we express it as a function of $p$: $f(p) = n p(1 - p) = n(p - p^2)$. Taking the derivative with respect to $p$ and setting it to zero: $f'(p) = n(1 - 2p) = 0$. This gives $1 - 2p = 0$,or $p = \frac{1}{2}$. Since $q = 1 - p$,we have $q = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting these values into the variance formula: $\sigma^2_{\max} = n 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{n}{4}$.

The maximum value of the variance of a Binomial distribution with parameters $n$ and $p$ is

Similar Questions

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. The mean of the number of queens is:

It is given that the discrete random variable is $X \sim B(n, p)$ and $P(X=2)=P(X=3)$. Find the mean of the distribution.

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. The mean of the number of kings is:

If $15$ coins are tossed,then the probability of getting $10$ heads is:

Numbers are selected at random,one at a time from the two-digit numbers $00, 01, 02, \dots, 99$ with replacement. An event $E$ occurs only if the product of the two digits of a selected number is $24$. If four numbers are selected,then the probability that the event $E$ occurs at least $3$ times is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module