amolent3u

2022-01-16

What are the mean and standard deviation of a binomial probability distribution with n=5 and $p=\frac{12}{17}$?

Tiefdruckot

Expert

Explanation:
Mean = np $=5\cdot \frac{12}{17}=\frac{60}{17}=3.52$
Standard deviation
$=\sqrt{npq}=\sqrt{np\left(1-p\right)}=\sqrt{5\left(\frac{12}{17}\right)\left(1-\frac{12}{17}\right)}=1.01$

Archie Jones

Expert

$\frac{12}{17}\approx 0.7$
$\mu =n\cdot p$
=5*0.7
=3.5
${\sigma }^{2}=np\left(1-p\right)$
=5*0.7(1-0.7)
=1.05
$\sigma =\sqrt{{\sigma }^{2}}=\sqrt{1.05}=1.01$

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