Binomial mean and standard deviation formulas

Mean and variance of Bernoulli distribution example

unfavorable = 40% ----> 0

favorable = 60% ----> 1

We map the values to a 0 and 1.

Mean:

$\mu = 0.4 \cdot 0 + 0.6 \cdot 1 = 0.6$

Variance:

$\sigma^2 = 0.4 \cdot (0-0.6)^2 + 0.6 \cdot (1 - 0.6)^2 \newline \sigma^2 = 0.4 \cdot 0.36 + 0.6 \cdot 0.16\newline \sigma^2 = 0.24\newline \sigma\ \ =\sqrt{0.24} = 0.49$

Bernoulli distribution mean and variance formulas

unfavorable = 40% ----> 0 ===> we change it to (1 - p) -> failure

favorable = 60% ----> 1 ===> we change it to (p) -> success

$\mu = (1-p) \cdot 0 + p \cdot 1 = p$

$\sigma^2=(1 -p)(0-p)^2 + p(1-p)^2\newline \sigma^2 = (1-p)p^2 +p(1-2p+p^2)\newline \sigma^2 = p^2 -p^3+p-2p^2+p^3\newline \sigma^2= p-p^2 \newline \sigma^2 = p(1-p)$

Expected value of a binomial variable

(see Statistics (hackerrank)/Poisson Distribution)

X = # of successes after $n$ trials where P(success) for each trial is $p$

$E(X)=n \cdot p$

$E(X+Y)=E(X)+E(Y)$

Finding the mean and standard deviation of a binomial random variable

A company produces cell phone chips. 2% of them are defect. A quality check involves randomly selecting and testing 500 chips.

What are the mean and standard deviation?

$X$= # of defective chips in 500 chip sample

$\mu_x=E(X)=n \cdot p \newline \mu_x = 500 \cdot 0.02 = 10$

$\sigma_x = \sqrt{\sigma_x^2} = \sqrt{Var(X)}=\sqrt{n \cdot p \cdot (1-p)}=\sqrt{500 \cdot 0.02 \cdot 0.98} = \sqrt{9.8} = 3.13$