expeditiupc

2022-01-17

What are the mean and standard deviation of a binomial probability distribution with n=20 and p=0.9?

Paul Mitchell

Mean 20
Standard deviation $\frac{3}{\sqrt{5}}$
Explanation:
$\mu =np$
=(20)*(0.9)
=18
$\sigma =\sqrt{np\left(1-p\right)}$
$=\sqrt{20\cdot \left(0.9\right)\cdot \left(0.1\right)}$
$=\frac{3}{\sqrt{5}}$
If the random variable X follows the binomial distribution with parameters $n\in \mathbb{N}$ and $p\in \left[0,1\right]$, tthe probability of getting exactly k successes in n trials, is given by
$Pr\left(X=k\right)=n{C}_{k}\cdot {p}^{k}\cdot {\left(1-p\right)}^{n-k}$
The mean is calculated from E(X).
The standard deviation comes from the square root of the variance,
$Var\left(X\right)=E\left({X}^{2}\right)-{\left[E\left(X\right)\right]}^{2}$
Where
$E\left(X\right)=\sum _{k=0}^{n}k\cdot Pr\left(X=k\right)$
$E\left({X}^{2}\right)=\sum _{k=0}^{n}{k}^{2}\cdot Pr\left(X=k\right)$

Jeffery Autrey

The mean and standard deviation of a binomial probability distribution with
n=200....(1)
p=0.3....(2)
can be determined as follows.
We know that the mean of a binomial probability distribution is given as:
$\mu =np$...(3)
And the standard deviation of a binomial probability distribution is given as:
$\sigma =\sqrt{np\left(1-p\right)}$...(4)
Substituting (1) and (2) in (3), we get:
$\mu =200\left(0.3\right)$
$\mu =60$
Substituting (1) and (2) in (4), we get:
$\sigma =\sqrt{200\left(0.3\right)\left(1-0.3\right)}$
$=\sqrt{200\left(0.3\right)\left(0.7\right)}$
$=\sqrt{42}$
$\sigma =6.48$

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