Mean, Variance and Standard Deviation

Mean and Variance of Binomial Distribution

If p is the probability of success and q is the probability of failure in a binomial trial, then the expected number of successes in n trials (i.e. the mean value of the binomial distribution) is

E(X) = μ = np

The variance of the binomial distribution is

V(X) = σ² = npq

Note: In a binomial distribution, only 2 parameters, namely n and p, are needed to determine the probability.

Example 1

A die is tossed $\displaystyle{3}$ times. What is the probability of

(a) No fives turning up?
(b) $\displaystyle{1}$ five?
(c) $\displaystyle{3}$ fives?

ANSWER:

 This is a binomial distribution because there are only 

$\displaystyle{2}$ possible outcomes (we get a $\displaystyle{5}$ or we don't).

Now, $\displaystyle{n}={3}$ for each part. Let $\displaystyle{X}=$ number of fives appearing.

(a) Here, x = 0.

$\displaystyle{P}{\left({X}={0}\right)}$ $\displaystyle={{C}_{{x}}^{{n}}}{p}^{x}{q}^{{{n}-{x}}}$ $\displaystyle={{C}_{{0}}^{{3}}}{\left(\frac{1}{{6}}\right)}^{0}{\left(\frac{5}{{6}}\right)}^{3}$ $\displaystyle=\frac{125}{{216}}$ $\displaystyle={0.5787}$

(b) Here, x = 1.

$\displaystyle{P}{\left({X}={1}\right)}$ $\displaystyle={{C}_{{x}}^{{n}}}{p}^{x}{q}^{{{n}-{x}}}$ $\displaystyle={{C}_{{1}}^{{3}}}{\left(\frac{1}{{6}}\right)}^{1}{\left(\frac{5}{{6}}\right)}^{2}$ $\displaystyle=\frac{75}{{216}}$ $\displaystyle={0.34722}$

(c) Here, x = 3.

$\displaystyle{P}{\left({X}={3}\right)}={{C}_{{x}}^{{n}}}{p}^{x}{q}^{{{n}-{x}}}={{C}_{{3}}^{{3}}}{\left(\frac{1}{{6}}\right)}^{3}{\left(\frac{5}{{6}}\right)}^{0}=\frac{1}{{216}}={4.6296}\times{10}^{ -{{3}}}$