Given p=0.3
Then q=1−p=1−0.3=0.7
By using the result P(y>y0)=qy0, we have
P(y>y0)=0.7y0 If P(y>y0)>0.1, then
0.7y0>0.1
Let 0.7y0=0.1
ln0.7y0=ln0.1
y0ln0.7=ln0.1
y0=ln0.7ln0.1≈6.4557
0.76=0.117649>0.1
0.77=0.0823543<0.1 The largest value of y is 6.
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