The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is

Solution :

Given that, for binomial distribution mean, np = 4 and variance, npq = 2.

\(\therefore\)  q = 1/2, but p + q = 1 \(\implies\) p = 1/2

and n \(\times\) \(1\over 2\) = 4 \(\implies\) n = 8

We know that,  P(X = r) = \(^nC_r p^r q^{n-r}\)

\(\therefore\)  P(X = 1) = \(^8C_1\)\(({1\over 2})^7\)\(({1\over 2})^1\)

= 1/32

Similar Questions

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new is

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is

In a series of 2n observations, half of them equals a and remaining half equal -a. If the standard deviation of the observation is 2, then |a| equal to

In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls?

Leave a Comment

Your email address will not be published.