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binomialcdf(7,.2,1) = ?

 Jan 14, 2016

Best Answer 

 #2
avatar+26364 
+10

binomialcdf( 7, 0.2, 1 ) = 0.5767168

 

\(\begin{array}{lcl} \boxed{~ \text{binomialcdf}(n,p,k) = \sum \limits_{i=0}^{k} \binom{n}{i} p^i(1-p)^{n-i} ~} \end{array} \\ \begin{array}{lcl} binomialcdf(7,0.2,1) &=& \sum \limits_{i=0}^{1} \binom{7}{i} 0.2^i(1-0.2)^{7-i} \\ &=& \sum \limits_{i=0}^{1} \binom{7}{i} 0.2^i\cdot 0.8^{7-i} \\ &=& \binom{7}{0} 0.2^0\cdot 0.8^{7-0}+\binom{7}{1} 0.2^1\cdot 0.8^{7-1} \\ &=& \binom{7}{0} 1\cdot 0.8^{7}+\binom{7}{1} 0.2^1\cdot 0.8^{6} \\ &=& \binom{7}{0} 0.8^{7}+\binom{7}{1} 0.2\cdot 0.8^{6} \qquad | \qquad \binom{7}{0} = 1 \qquad \binom{7}{1} = 7\\ &=& 1\cdot 0.8^{7}+7\cdot 0.2\cdot 0.8^{6}\\ &=& 0.8^{7}+1.4\cdot 0.8^{6}\\ &=& 0.2097152+ 0.3670016 \\ &=& 0.5767168 \\ &=& 57.67\ \% \end{array} \)

 

laugh

 Jan 14, 2016
edited by heureka  Jan 14, 2016
edited by heureka  Jan 14, 2016
 #1
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tbh idgaf.

 Jan 14, 2016
 #2
avatar+26364 
+10
Best Answer

binomialcdf( 7, 0.2, 1 ) = 0.5767168

 

\(\begin{array}{lcl} \boxed{~ \text{binomialcdf}(n,p,k) = \sum \limits_{i=0}^{k} \binom{n}{i} p^i(1-p)^{n-i} ~} \end{array} \\ \begin{array}{lcl} binomialcdf(7,0.2,1) &=& \sum \limits_{i=0}^{1} \binom{7}{i} 0.2^i(1-0.2)^{7-i} \\ &=& \sum \limits_{i=0}^{1} \binom{7}{i} 0.2^i\cdot 0.8^{7-i} \\ &=& \binom{7}{0} 0.2^0\cdot 0.8^{7-0}+\binom{7}{1} 0.2^1\cdot 0.8^{7-1} \\ &=& \binom{7}{0} 1\cdot 0.8^{7}+\binom{7}{1} 0.2^1\cdot 0.8^{6} \\ &=& \binom{7}{0} 0.8^{7}+\binom{7}{1} 0.2\cdot 0.8^{6} \qquad | \qquad \binom{7}{0} = 1 \qquad \binom{7}{1} = 7\\ &=& 1\cdot 0.8^{7}+7\cdot 0.2\cdot 0.8^{6}\\ &=& 0.8^{7}+1.4\cdot 0.8^{6}\\ &=& 0.2097152+ 0.3670016 \\ &=& 0.5767168 \\ &=& 57.67\ \% \end{array} \)

 

laugh

heureka Jan 14, 2016
edited by heureka  Jan 14, 2016
edited by heureka  Jan 14, 2016

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