Test Index

BITSAT 2018 Paper

Section: Mathematics

Question : 122 of 150

Marks: +1, -0

In a binomial distribution, the mean is 4 and variance is 3. Then its mode is :

Solution:

In Binomial distribution, Mean = np,Variance = npq and the mode is r if for x = r, the probability function p(x) is maximum.Given np = 4 and npq = 3

∴ q =

\frac{3}{4}

and p = 1 - q = 1 -

\frac{3}{4}

\frac{1}{4}

Also, n =

\frac{4}{p}

\frac{4}{\frac{1}{4}}

= 16

Now, (n + 1) p = (16 + 1)

\frac{1}{4}

\frac{17}{4}

= 4 +

\frac{1}{4}

⇒ The distribution will have unique mode(unimodal) & the mode = 4

Go to Question:

Prev Question

Next Question