Sapewa

Answered

2022-01-18

What are the mean and standard deviation of a binomial probability distribution with n=777 and $p=\frac{19}{21}$?

Answer & Explanation

otoplilp1

Expert

2022-01-18Added 41 answers

Explanation:
$Mean=np=777\cdot \frac{19}{21}=703$
$Variance=np\left(1-p\right)=777\cdot \frac{19}{21}\cdot \frac{2}{21}=66.9$

boronganfh

Expert

2022-01-19Added 33 answers

$\frac{19}{21}\approx 0.9$
$\mu =n\cdot p$
=777*0.9
=699.3
${\sigma }^{2}=np\left(1-p\right)$
=777*0.9(1-0.9)
=69.93
$\sigma =\sqrt{{\sigma }^{2}}=\sqrt{69.93}=8.18$

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