What are the mean and standard deviation of a binomial probability distribution with $n = 13$ $n=13$ and $p = \frac{5}{24}$ $p=5/24$ ?

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What are the mean and standard deviation of a binomial probability distribution with $n = 13$ $n=13$ and $p = \frac{5}{24}$ $p=5/24$ ?

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Answer 1 · 2016-03-26T05:48:57

Mean is $2.708$ $2.708$ and Standard Deviation is $1.464$ $1.464$

Explanation:

Binomial distribution is a frequency distribution of possible number of successful outcomes in given number of trials $n$ $n$ , in each of which there is the probability of success is $p$ $p$ (and $q = 1 - p$ $q=1-p$ is probability of failure).

The mean and standard deviation of a binomial distribution are given by $n p$ $np$ and $\sqrt{n p q}$ $sqrt(npq)$ .

Here $n = 13$ $n=13$ , $p = \frac{5}{24}$ $p=5/24$ and $q = 1 - \frac{5}{24} = \frac{19}{24}$ $q=1-5/24=19/24$ .

Hence Mean is $13 \times \frac{5}{24} = 2.708$ $13xx5/24=2.708$

Standard Deviation is $\sqrt{13 \times \frac{5}{24} \times \frac{19}{24}}$ $sqrt(13xx5/24xx19/24)$

= $\frac{1}{24} \times \sqrt{1235} = \sqrt{4.8696} = 1.464$ $1/24xxsqrt1235=sqrt4.8696=1.464$

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What are the mean and standard deviation of a binomial probability distribution with $n = 13$ $n=13$ and $p = \frac{5}{24}$ $p=5/24$ ?

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Explanation:

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