Find mean, variance, and standard deviation for a Binomial distribution

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asked Nov 14, 2023 58.3k views

1 Answer

Ted Benson · Answer 1 · 2023-11-21T00:54:52+0000

The mean, variance, and standard deviation of a binomial distribution with parameters n and p are:

$\begin{gathered} \operatorname{mean}=n\cdot p \\ \text{variance = n}\cdot p\cdot(1-p) \\ \text{ Standard deviation =}\sqrt[]{n\cdot p\cdot(1-p)} \end{gathered}$

So, replacing the values of n and p, we get:

$\begin{gathered} \operatorname{mean}=128\cdot0.49=62.72 \\ \text{variance}=128\cdot0.49\cdot(1-0.49)=31.9872 \\ \text{ standard deviation = }\sqrt[]{31.9872}=5.6557 \end{gathered}$

Answer: mean = 62.72

variance = 31.9872

standard deviation = 5.6557

Find mean, variance, and standard deviation for a Binomial distribution

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