Define the steps to read a cumulative binomial probability table.

Hint:A binomial probability refers to the probability of getting EXACTLY r successes in an exceedingly specific number of trials. Cumulative binomial probability refers to the probability that the worth of a binomial random variable falls within a specified range. Cumulative binomial probability tables are used to find \[P(X \leqslant x)\]for the distribution\[X-B(n,p)\]

Complete step by step solution:
We know that cumulative binomial probability tables are used to find \[P(X \leqslant x)\]for the distribution\[X-B(n,p)\].
Basic rules that are used to find probabilities in a binomial distribution table:
\[P(X < x) = P(X \leqslant x - 1)\]…….(1)
\[P(X \geqslant x) = 1 - P(X \leqslant x - 1)\]…….(2)
\[P(X > x) = 1 - P(X \leqslant x)\]………(3)
\[P(A < X \leqslant B) = P(X \leqslant B) - P(X \leqslant A)\]……..(4)
There is a separate table for every sample size .First step is to find the proper table of
n (where n is equal to your sample size).
Then find the column on the same table with the probability of your distribution. The number in the row \[x = a\]
For example, to find \[P(4 \leqslant X \leqslant 9)\]for the distribution \[X-B(14,0.55)\], head to the table for \[n = 14\].
Then find the column \[p = 0.55\]. Seek for the row \[x = 9\]in this column, which
supplies 0.8328, then searches for\[x = 4\] in this column, which provides 0.0114. (Here we have assumed the values of the probabilities)
Thus using the formula (4) from the above list of formulas we’ll get, \[P(4 \leqslant X \leqslant 9) = P(X \leqslant 9) - P(X \leqslant 4) = 0.8328 - 0.0114 = 0.8214\]

Note: The binomial cumulative distribution function enables you to obtain the probability of observing less or equal to x successes in n trials, with the probability p of success on one trial. The binomial cumulative distribution function for a given value x and a given pair of parameters n and p is \[y = F(x|n,p) = \sum\limits_{i = 0}^x {\left( {\begin{array}{*{20}{c}}
n \\
 i
\end{array}} \right){p^i}{{(1 - p)}^{(n - i)}}{I_{(0,1,...,n)}}(i)} \]