Why do we have to calculate 1-binomcdf for P(X>11)?

Question

Can't we just calculate binomcdf with left and right bound like the normalcdf?

Who says we have to? I have never done so in my life. Where does the question come from? — Ross Millikan, Jul 26 '18 at 14:43
@callculus Yes, sure. I use this formula for a long time bu do not really understand by we havo to do 1-binomcdf and this is not needed for normalcdf... — WinstonCherf, Jul 26 '18 at 14:58
@WinstonCherf If you use the cdf-function (table or calculator) then you have to apply the converse probability for greater relations ($P(X>k)=$), in both cases (normal dist and binomial dist). — callculus42, Jul 26 '18 at 15:01
@callculus See example 2 with the normalcdf... we don't use 1-normalcdf... — WinstonCherf, Jul 26 '18 at 15:05
@WinstonCherf We need to know how the whole example looks like. Without further information we cannot give a informative answer. The first exercise is not important anymore for your question. — callculus42, Jul 26 '18 at 15:11
@callculus In our textbook we looked at P(X>k) for normal end binomial distibutions. We saw that for the binomial distribution, you have to do 1-P(X<=k) (example 1) and for normal distribution, you just can just calculate P(X>k) (example 2). Why can't we just calculate P(X>k) for the binomial in example 1? — WinstonCherf, Jul 26 '18 at 15:15
@WinstonCherf In your case the function normalcdf is not the statistical cdf. In your case you can input the lower bound and upper bound, which are $0$ and $10^{99}$. This es equal to $P(V>0)$. — callculus42, Jul 26 '18 at 15:21
@callculus yes indeed! Is this also possible for the binomcdf? — WinstonCherf, Jul 26 '18 at 15:32
@WinstonCherf This depends on the software you use. Do you know which software is supposed to use in your course? — callculus42, Jul 26 '18 at 15:36
@WinstonCherf Maybe, try binomcdf(n,p,lower bound, upper bound). In your example binomcdf(12,0.25,4, 12) — callculus42, Jul 26 '18 at 15:46
@callculus Yes it works! :) But you said first "If you use the cdf-function then you have to apply the converse probability for greater relations (P(X>k)=) in both cases (normal dist and binomial dist). In your case the function normalcdf is not the statistical cdf." What do you mean by that? — WinstonCherf, Jul 26 '18 at 20:59
@WinstonCherf Usually the cdf is defined for $P(X\leq x)$. See here.It cannot be restricted by a lower bound. If you want to calculate $P(X\geq x)$ you can use the cdf. But you have to apply the converse probability. If $X$ is a continuous random variable the equation is $P(X\leq X)=1-P(X\leq x)$. If $X$ is a discrete random variable the equation is $P(X\leq X)=1-P(X\leq x-1)$. — callculus42, Jul 26 '18 at 21:40
@callculus Thnx! Clear, I just think you mean P(X≥x)=1−P(X≤x) for continious and P(X≥x)=1−P(X≤x−1) for discrete. You used P(X≤x) — WinstonCherf, Jul 26 '18 at 22:54
@WinstonCherf Sure, that´s what I wanted to type. I don´t know why I Have made this typos. If we see it positive it was a good training for you. — callculus42, Jul 27 '18 at 01:54

BruceET · Answer 1 · 2018-07-27T04:24:59.650

You have a random variable $X \sim \mathsf{Binom}(n = 12,\,p = 1/4)$ and you seek $P(X > 11).$ There are many ways to approach this problem (with or without whatever software uses the function binomcdf). I will show three main approaches. I think the first is best because it is so simple. But I guess the author may have expected you to use the last.

(a) Using the binomial PDF formula: $$P(X > 11) = P(X = 12) = {12 \choose 12}(1/4)^{12}(1 - 1/4)^0 = (1/4)^{12} = 5.9605 \times 10^{-08}.$$

Or in R statistical software, where dbinom is a binomial PDF:

dbinom(12, 12, .25)
## 5.960464e-08

(b) Using the binomial CDF: This would be tedious to compute by hand, but some books have limited binomial CDF tables, and this choice of parameters may be included there.

$$P(X > 11) = 1 - P(X \le 11) = 1 - [P(X=0) + P(X=1) + \cdots + P(X = 11)].$$

In R, where pbinom is a binomial CDF, the syntax is somewhat similar to what you showed in your question.

1 - pbinom(11, 12, 1/4)
## 5.960464e-08

(c) Using normal approximation: Some texts may ask you to evaluate this using a normal approximation to the binomial, but that approximation will work rather poorly for this problem because $n=12$ is relatively small and $p = 1/4$ is relatively far from $1/2.$

Here is how the normal approximation would be attempted. First, $\mu = E(X) = np = 12(1/4) = 3.$ Also, $\sigma = SD(X) = \sqrt{np(1-p)} = \sqrt{9/4} = 3/2.$ Then $$P(X > 11) = P(X > 11.5) = P\left(\frac{X - \mu}{\sigma} > \frac{11.5-3}{1.5}\right) \approx P(Z > 5.67) \approx 0,$$ where $Z$ has the standard normal distribution, which places almost no probabiltiy above 5.67. The normal approximation "works" only because the result is so tiny that the mis-match is invisible.

1 - pnorm(5.67)
## 7.139876e-09

In the figure below the exact binomial probabilities are shown as vertical blue bars and the (ill-fitting) PDF of the normal approximation is shown as a black curve. The probability of interest is to the right of the vertical broken red line; this is a region where neither the binomial nor the normal distribution has appreciable probability.

Note: I spent time with the normal approximation because the problem seems to have been contrived so that $\mu$ and $\sigma$ are very easy to calculate. This makes me suspect that the normal approximation may be the intended method.

Why do we have to calculate 1-binomcdf for P(X>11)?

1 Answers1