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

Solution :

Given that, mean = 4 \(\implies\) np = 4

And Variance = 2 \(\implies\) npq = 2 \(\implies\) 4q = 2

\(\implies\)  q = \(1\over 2\)

\(\therefore\)   p = 1 – q = 1 – \(1\over 2\) = \(1\over 2\)

Also, n = 8

Probability of 2 successes = P(X = 2) = \(^8C_2\)\(p^2\)\(q^6\)

= \(8!\over {2!\times 6!}\) \(\times\) \(({1\over 2})^2\) \(\times\) \(({1\over 2})^2\)

= \(28\over 256\)


Similar Questions

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

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

Suppose a population A has 100 observation 101, 102, ….. , 200 and another population B has 100 observations 151, 152, …. , 250. If \(V_A\) and \(V_B\) represent the variance of the two populations respectively, then \(V_A\over V_B\) 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Your email address will not be published.