Binomial Distribution Solving for N I am having trouble trying to help a sibling do their statistics homework, so I thought someone could help me. The problem asks to solve for N here:
$$\sum_{i=6}^N\binom Ni(0.25)^i(0.75)^{N-i}<0.005$$
Could someone please explain how to solve this by hand, as well as how to input this into a TI-83 or 84 calculator? If you could, please explain thoroughly, as it has been quite a while since I have taken statistics. Thank you!
 A: The question asks for $N$ such that $A=P(\operatorname B(N,0.25)\ge6)<0.005$.
Immediately we have $N\ge6$. Now note that $A$ increases with $N$, simply because the bulk of the binomial distribution is shifted farther and farther away from the fixed bound of 6. Hence we may start from $N=6$ and increment, calculating $A$ each time and stopping when the sum becomes greater than 0.005:
$$P(\operatorname B(6,0.25)\ge6)=0.000244<0.005$$
$$P(\operatorname B(7,0.25)\ge6)=0.00134<0.005$$
$$P(\operatorname B(8,0.25)\ge6)=0.00423<0.005$$
$$P(\operatorname B(9,0.25)\ge6)=0.00999>0.005$$
Hence $N=6,7,8$. I do not have a TI-83 or TI-84 on hand (I have an Nspire, and even then I only use it when absolutely required), but you should be able to look up the functions for the binomial coefficient and summation and then solve for $N$ with the procedure I've outlined above.
A: Comment.  Verifying @ParclyTaxel's answer [which I have up-voted (+1), and for which you may also want to click the check mark to 'accept'], using R statistical software:
the solution is $N \le 8.$
 N = 6:15;  prob = 1 - pbinom(5, N, 1/4)
 plot(N, prob)
 abline(h = .005)
 cbind(N, prob)  # ignore numbers in brackets
        N         prob
  [1,]  6 0.0002441406
  [2,]  7 0.0013427734
  [3,]  8 0.0042266846
  [4,]  9 0.0099945068
  [5,] 10 0.0197277069
  [6,] 11 0.0343275070
  [7,] 12 0.0544022322
  [8,] 13 0.0802125931
  [9,] 14 0.1116689704
 [10,] 15 0.1483680774


It is possible that this is just a drill problem for practice using a statistical calculator . But it might be possible to solve this using normal approximations, even though
the the $N$s are a little too small for really accurate answers. You might check
to see if normal approximation to binomial probabilities is covered
in the text just before this problem.
If $N = 8,$ then $X \sim Binom(8, 1/4)$ with $\mu = E(X) = Np = N/4 = 2$ and
$\sigma = SD(X) = \sqrt{Np(1-p)} = \sqrt{3N/16} = 1.225.$ So 
$$P(X \ge 6) = P(X > 5.5) = 
P\left(\frac{X - \mu}{\sigma} > \frac{5.5 - 2}{1.225} = 2.86\right)
\approx P(Z > 2.86) \approx 0.0021,$$
where $Z$ is standard normal. The final (approximate) answer
is from printed tables or software.
Then try to the same thing for $N = 9$ and see if the answer exceeds 0.005.
