What are the mean and standard deviation of a binomial probability distribution with $n = 130$ $n=130$ and $p = \frac{3}{4}$ $p=3/4$ ?

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What are the mean and standard deviation of a binomial probability distribution with $n = 130$ $n=130$ and $p = \frac{3}{4}$ $p=3/4$ ?

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Answer 1 · 2015-11-26T13:58:35

Mean is nxp and variance is nxpxq

Explanation:

Here n = 130

p = $\frac{3}{4}$ $3/4$
q = $\frac{1}{4}$ $1/4$
So mean = 130 x $(\frac{3}{4})$ $(3/4)$ = $\frac{390}{4}$ $390 / 4$ = 97.5
Variance = 130 x $\frac{3}{4}$ $3/4$ x $\frac{1}{4}$ $1/4$ = $\frac{390}{16}$ $390/16$ = 24.375
Standard deviation = $\sqrt{v a r i a n c e}$ $sqrt(variance)$ = $\sqrt{24.375}$ $sqrt(24.375)$

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What are the mean and standard deviation of a binomial probability distribution with $n = 130$ $n=130$ and $p = \frac{3}{4}$ $p=3/4$ ?

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What are the mean and standard deviation of a binomial probability distribution with n=130 n=130n=130 and p=3/4 p=34p=3/4 ?

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What are the mean and standard deviation of a binomial probability distribution with $n = 130$ $n=130$ and $p = \frac{3}{4}$ $p=3/4$ ?