Expert Answer to "What are the mean and standard deviation of..."

zgribestika

Answered question

2022-01-19

What are the mean and standard deviation of a binomial probability distribution with n=94 and

p = \frac{32}{37}

Buck Henry

Beginner2022-01-19Added 33 answers

Explanation:
Mean of a binomial probability distribution is given by np and standard deviation is given by

\sqrt{n} p q

, where q=1-p.
In the given example,

n p = 94 \cdot \frac{32}{37} = 81.2

q = 1 - \frac{32}{37} = \frac{5}{37}, \sqrt{n} p q = \sqrt{94 \cdot \frac{32}{37} \cdot \frac{5}{37}} = \frac{4}{37} \cdot \sqrt{940} = 3.3

Hence mean is 81.2 and standard deviation is 3.3

intacte87

Beginner2022-01-20Added 42 answers

The formula for the population mean is \mu = n*p, so therefore we get:

μ = n \cdot p

= 94 \cdot \frac{32}{37}

=81.2
On the other hand, the variance is computed using the following formula:

σ^{2} = n p (1 - p)

. Therefore, the calculation goes like:

σ^{2} = n p (1 - p)

= 94 \cdot \frac{32}{37} (1 - \frac{32}{37})

=11.3

σ = \sqrt{σ^{2}} = \sqrt{11.3} = 3.3

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