Binomial Distribution Question

**db5vry** · May 24th 2009, 03:27 PM

The random variable X has the distribution B (n, 0.1). Given that the mean and standard deviation of X are equal, find the value of n. [5]

The only things I could work out were:

E(X)(mean) = n x 0.1
SD(X) = root of VAR(X)
VAR(X) = n x 0.1 x 0.9
VAR(X) = n x 0.09

n x 0.1 = root of 0.09 x n

What happens from here? Is it meant to be solved in a different way?

**noob mathematician** · May 24th 2009, 04:50 PM

Originally Posted by db5vry

The random variable X has the distribution B (n, 0.1). Given that the mean and standard deviation of X are equal, find the value of n. [5]

The only things I could work out were:

E(X)(mean) = n x 0.1
SD(X) = root of VAR(X)
VAR(X) = n x 0.1 x 0.9
VAR(X) = n x 0.09

n x 0.1 = root of 0.09 x n

What happens from here? Is it meant to be solved in a different way?

Guess you are on the right track.
square your equation to get:

$0.01n^2=0.09n$

$0.01n^2-0.09n=0$

$n(0.01n-0.09)=0$

$\text{n=0 (N.A) or n=9}$

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