The expected value np=6
Variance npq=2.4
q="\\frac{npq} {np} =\\frac {2.4}{6}" =0.4
p=1-0.4=0.6
np=6=n*0.6
n="\\frac {6}{0.6}" =10
P(X=5)
"P(X=x) =\\binom {n} {x} p^x (1-p)^{n-x}"
"P(X=5) =\\binom {10}{5}*0.6^5 *0.4^5"
=0.2007
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