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

Beginner2022-01-17Added 40 answers

Mean 20

Standard deviation$\frac{3}{\sqrt{5}}$

Explanation:

$\mu =np$

=(20)*(0.9)

=18

$\sigma =\sqrt{np(1-p)}$

$=\sqrt{20\cdot \left(0.9\right)\cdot \left(0.1\right)}$

$=\frac{3}{\sqrt{5}}$

If the random variable X follows the binomial distribution with parameters$n\in \mathbb{N}$ and $p\in [0,1]$ , tthe probability of getting exactly k successes in n trials, is given by

$Pr(X=k)=n{C}_{k}\cdot {p}^{k}\cdot {(1-p)}^{n-k}$

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

$E\left(X\right)=\sum _{k=0}^{n}k\cdot Pr(X=k)$

$E\left({X}^{2}\right)=\sum _{k=0}^{n}{k}^{2}\cdot Pr(X=k)$

Standard deviation

Explanation:

=(20)*(0.9)

=18

If the random variable X follows the binomial distribution with parameters

The mean is calculated from E(X).

The standard deviation comes from the square root of the variance,

Where

Jeffery Autrey

Beginner2022-01-18Added 35 answers

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:

$\sigma =\sqrt{np(1-p)}$ ...(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:

$\sigma =\sqrt{200\left(0.3\right)(1-0.3)}$

$=\sqrt{200\left(0.3\right)\left(0.7\right)}$

$=\sqrt{42}$

$\sigma =6.48$

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:

And the standard deviation of a binomial probability distribution is given as:

Substituting (1) and (2) in (3), we get:

Substituting (1) and (2) in (4), we get:

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