Binomial distribution regarding: (≤, >, etc.)

[iamlorde]

Question is as follows:

Let X be a binomial random variable with p = 0 2 .
and n = 20. Use the binomial table in Appendix A to determine
the following probabilities.
(a) P(X ≤ 3) (b) P(X > 10)
(c) P(X = 6) (d) P(6 ≤ X ≤ 11)

(a) = 0.4114 is the answer. Yet all I see from this answer is that X is simple equal to "0.4114". If it is "X ≤ 3" shouldn't "0.2061", "0.0692", and "0.0115" contribute to the answer somehow because they are "<" smaller than 3?

I feel like I may be missing a fundamental element here. How do I proceed with these in general? My logic seems flawed on this matter.

For example: (c) = 0.9133-0.8042=0.1091; how is this possible. Why isn't this straight out "0.9133", the value directly next to #6?

Please enlighten me on this matter. Thank you.

 

[MarkFL]

The numbers in the column of your table are cumulative, and so for a) you would simply read from the table to get:

a) $$P(x\le3)=0.4114$$

And for c), you would do the following:

c) $$P(x=6)=P(x\le6)-P(x\le5)=0.9133-0.8042=0.1091$$

How do you now suppose you would do parts b) and d)?

 

[iamlorde]

So, I would think,

(b) P(X > 10) = [!not](everything up to and including 10) = 1-0.9994 = 0.0006

(d) P(6 ≤ X ≤ 11) :

X is between 6 and 11, AND it includes them both, so I choose one above 11, so 12 which is: 1.0000. Then I choose one below 6, so 5, so that I can include 6. Hence: 0.8042.

Now I subtract: 1 - 0.8042 = 0.1958 (YET THIS IS FALSE)
My solution says: 0.1957?

 

[Siron]

Your solution for (d) is almost correct. The problem is that you've calculated $\mathbb{P}(6 \leq X \leq 12)$ because you've also included $12$. We have

$$\mathbb{P}(6 \leq X \leq 11) = \mathbb{P}(X \in \{6,7,8,9,10,11\}) = \mathbb{P}(X \leq 11) - \mathbb{P}(X \leq 5) = 0.1957$$.