What are the mean and standard deviation of a binomial

Question

What are the mean and standard deviation of a binomial

expeditiupc 2022-01-17 Answered

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

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Expert Answer

Paul Mitchell

Answered 2022-01-17 Author has 40 answers

Mean 20
Standard deviation

\frac{3}{\sqrt{5}}

Explanation:

μ = n p

=(20)*(0.9)
=18

σ = \sqrt{n p (1 - p)}

= \sqrt{20 \cdot (0.9) \cdot (0.1)}

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

If the random variable X follows the binomial distribution with parameters

n \in N

and

p \in [0, 1]

, tthe probability of getting exactly k successes in n trials, is given by

P r (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,

V a r (X) = E (X^{2}) - {[E (X)]}^{2}

Where

E (X) = \sum_{k = 0}^{n} k \cdot P r (X = k)

E (X^{2}) = \sum_{k = 0}^{n} k^{2} \cdot P r (X = k)

This is helpful 0

Jeffery Autrey

Answered 2022-01-18 Author has 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:

μ = n p

...(3)
And the standard deviation of a binomial probability distribution is given as:

σ = \sqrt{n p (1 - p)}

...(4)
Substituting (1) and (2) in (3), we get:

μ = 200 (0.3)

μ = 60

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

σ = \sqrt{200 (0.3) (1 - 0.3)}

= \sqrt{200 (0.3) (0.7)}

= \sqrt{42}

σ = 6.48

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