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) = [tex]^nC_r p^r q^n^-^r = ^{12}C_0[/tex] [tex]0.7^0[/tex] [tex]0.3^{12} = 1\times1\times0.0053 = 0.0053[/tex]

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.

 

[wais]

The smallest value of d can be calculated by using the cumulative distribution function (CDF) of the binomial distribution. The CDF represents the probability that the random variable D is less than or equal to a certain value. In this case, we are interested in finding the smallest value of d such that the probability of D being greater than d is less than 0.90.

We can use a calculator or software to find the CDF for different values of d and then determine the smallest value that satisfies the condition. Alternatively, we can use a table of values for the binomial distribution to find the smallest value of d.

For example, using a calculator, we can find that P(D≤7) = 0.8862 and P(D≤8) = 0.9574. This means that the smallest value of d for which P(D>d) <0.90 is d=8.

Using a table, we can find the value of d by looking for the first row where the cumulative probability (in this case, 0.9574) is greater than 0.90. This corresponds to d=8.

In summary, the smallest value of d for which P(D>d) <0.90 is d=8, meaning that there is a 90% chance that D will be less than or equal to 8. I hope this helps to shed some light on the dilemma.