What are the mean and standard deviation of a binomial probability distribution with $n = 130$ $n=130$ and $p = \frac{7}{9}$ $p=7/9$ ?

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Answer 1 · 2016-05-07T17:18:11

Mean = $101 \frac{1}{9}$ $101 1/9$
Standard Deviation = 4.74 (3 sig. fig.)

$X ~ B (130, \frac{7}{9})$ $X~B(130,7/9)$

Mean = np, so $130 \times \frac{7}{9} = 101 \frac{1}{9}$ $130xx7/9=101 1/9$

Variance = npq.

We know that $p + q = 1$ $p + q = 1$
So $q = \frac{2}{9}$ $q=2/9$

Therefore the variance is
$130 \times \frac{7}{9} \times \frac{2}{9} = 22 \frac{38}{81}$ $130xx7/9xx2/9=22 38/81$

Standard deviation is the square root of the variance, so s.d. = $\sqrt{22 \frac{38}{81}} = 4.74$ $sqrt(22 38/81) = 4.74$ (3 sig. fig.)

What are the mean and standard deviation of a binomial probability distribution with n=130n=130 and p=79p=7/9 ?