If X is a binomial variate with mean frac{16} | vedclass

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

Question

(A) For a binomial distribution,mean $\mu = np = \frac{16}{5}$ and variance $\sigma^2 = npq = \frac{48}{25}$.Dividing variance by mean,we get $q = \fr

Vedclass · Accepted Answer

$\frac{3^6(169)}{5^8}$

Vedclass · Answer

(A) For a binomial distribution,mean $\mu = np = \frac{16}{5}$ and variance $\sigma^2 = npq = \frac{48}{25}$.Dividing variance by mean,we get $q = \frac{npq}{np} = \frac{48/25}{16/5} = \frac{48}{25} 	imes \frac{5}{16} = \frac{3}{5}$.Since $p + q = 1$,we have $p = 1 - \frac{3}{5} = \frac{2}{5}$.Substituting $p$ in $np = \frac{16}{5}$,we get $n 	imes \frac{2}{5} = \frac{16}{5}$,so $n = 8$.The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k} = \binom{8}{k} (\frac{2}{5})^k (\frac{3}{5})^{8-k}$.We need $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.$P(X=0) = \binom{8}{0} (\frac{2}{5})^0 (\frac{3}{5})^8 = \frac{3^8}{5^8}$.$P(X=1) = \binom{8}{1} (\frac{2}{5})^1 (\frac{3}{5})^7 = 8 	imes \frac{2}{5} 	imes \frac{3^7}{5^7} = \frac{16 	imes 3^7}{5^8} = \frac{16 	imes 3 	imes 3^6}{5^8} = \frac{48 	imes 3^6}{5^8}$.$P(X=2) = \binom{8}{2} (\frac{2}{5})^2 (\frac{3}{5})^6 = 28 	imes \frac{4}{25} 	imes \frac{3^6}{5^6} = \frac{112 	imes 3^6}{5^8}$.Summing these: $P(X \leq 2) = \frac{3^8 + 48 	imes 3^6 + 112 	imes 3^6}{5^8} = \frac{9 	imes 3^6 + 160 	imes 3^6}{5^8} = \frac{169 	imes 3^6}{5^8}$.

If $X$ is a binomial variate with mean $\frac{16}{5}$ and variance $\frac{48}{25}$,then $P(X \leq 2) = $

Similar Questions

The mean and variance of a binomial distribution are $6$ and $4$. The parameter $n$ is

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

If the sum and the product of the mean and variance of a binomial distribution are $24$ and $128$ respectively,then the probability of one or two successes is:

Two cards are drawn at random one after the other with replacement from a pack of $52$ playing cards. Then the variance of the random variable of the number of spade cards among the drawn cards is

Vedclass Test Series

Exam Paper Generator

Online Exam Module