Answer to Question #195499 in Statistics and Probability for Mohammed Moro

Question #195499

If YY has a geometric distribution with success probability 0.3,what is the largest value of yy,such that P9Y>y)>0.1P9Y>y)>0.1 ?


1
Expert's answer
2021-05-20T08:30:09-0400

Given p=0.3p=0.3

Then q=1p=10.3=0.7q=1-p=1-0.3=0.7

By using the result P(y>y0)=qy0P(y>y_0)=q^{y_0}, we have


P(y>y0)=0.7y0P(y>y_0)=0.7^{y_0}

If P(y>y0)>0.1,P(y>y_0)>0.1, then


0.7y0>0.10.7^{y_0}>0.1

Let 0.7y0=0.10.7^{y_0}=0.1

ln0.7y0=ln0.1\ln{0.7^{y_0}}=\ln{0.1}

y0ln0.7=ln0.1y_0\ln{0.7}=\ln{0.1}

y0=ln0.1ln0.76.4557y_0=\dfrac{\ln 0.1}{\ln 0.7}\approx6.4557

0.76=0.117649>0.10.7^{6}=0.117649>0.1

0.77=0.0823543<0.10.7^{7}=0.0823543<0.1

The largest value of yy is 6.



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