What are the mean and standard deviation of a binomial probability distribution with n=8 and...

Talamancoeb

Answered

2022-01-18

What are the mean and standard deviation of a binomial probability distribution with n=8 and $p=\frac{7}{29}$?

Answer & Explanation

psor32

Expert

2022-01-18Added 33 answers

$mean=\frac{56}{29}$ $s\mathrm{tan}dard\text{}deviation=\frac{4\sqrt{77}}{29}$
Explanation:
Given Binomial Probability Distribution
$mean=np=8\cdot \frac{7}{29}=\frac{56}{29}$ $s\mathrm{tan}dard\text{}deviation=\sqrt{np(1-p)}=\sqrt{8\cdot \frac{7}{29}\cdot \frac{22}{29}}=\frac{4\sqrt{77}}{29}$

Thomas White

Expert

2022-01-19Added 40 answers

$\mu =n\cdot p$ $=8\cdot \frac{7}{29}$
=1.92
On the other hand, the variance is computed using the following formula: ${\sigma}^{2}=np(1-p)$. Therefore, the calculation goes like:
${\sigma}^{2}=np(1-p)$ $=8\cdot \frac{7}{29}(1-\frac{7}{29})$
=1.4592
And finally, taking square root to the variance we get that the population standard deviation is:
$\sigma =\sqrt{{\sigma}^{2}}=\sqrt{1.4592}=1.208$