Binomial distribution smallest value

[bob4000]

could someone please shed some light upon the following dilemma:

Given that D~B(12,0.7), calculate the smallest value of d such that
P(D>d) <0.90.

much obliged

 

[The Bob]

could someone please shed some light upon the following dilemma:

Given that D~B(12,0.7), calculate the smallest value of d such that
P(D>d) <0.90.

I am going to replace the D with an X and the d with an x and the B with a Bin.

X~Bin(12,0.7)
From this you know that n=12, p=0.7, q=0.3 and x=d.
n is the number of trials, p is the probability of the event n happening and q is 1 - p.

So X~Bin(12,0.7) = $$^nC_r p^r q^n^-^r = ^{12}C_0$$ $$0.7^0$$ $$0.3^{12} = 1\times1\times0.0053 = 0.0053$$

All you then need to do this for the next few until you get an answer which answers your question.

The Bob (2004 ©)

EDIT: I think this post needs a little more information but I have not the time now. Sorry.