logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

The mean of a binomial distribution is $5$ and its standard deviation is $2.$ Then the value of $n$ and $p$ are

\[\begin{array}{1 1}(1)\left(\frac{4}{5},25\right)&(2)\left(25,\frac{4}{5}\right)\\(3)\left(\frac{1}{5},25\right)&(4)\left(25,\frac{1}{5}\right)\end{array}\]

Can you answer this question?
 
 

1 Answer

0 votes
Given mean $np= 5$
Standard deviation $\sqrt {npq} =2$
$npq= 4$
$\large\frac{npq}{np}= \frac{4}{5} $
$q= \large\frac{4}{5}$
$p=1-q=1-\large\frac{4}{5}=\large\frac{1}{5}$
$np=5=n \times \large\frac{1}{5} $$=5$
$n=25$
$(n,p)$ is $(25,\large\frac{1}{5})$
Hence 4 is the correct answer.
answered May 23, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...