Question
If Y gas a geometric distribution with success probability 0.3,what is the largest value of y such that P(Y>y)≥0.1
I had my answer to be 6.But I'm not sure
I had my answer to be 6.But I'm not sure
Answers
Gabriel
P=0.3
q=1-p =0.7
P(Y=y)=q^y=0.7^y
y such that 0.7^y=0.1
I apply ln to both sides and I get y to be 6
But I'm not sure with my answer
q=1-p =0.7
P(Y=y)=q^y=0.7^y
y such that 0.7^y=0.1
I apply ln to both sides and I get y to be 6
But I'm not sure with my answer