Download Statistics 211/Labpack/Binomial Distribution * P(X = x) = n x Ê Ë Á ˆ

Statistics 211/Labpack/Binomial Distribution
Binomial:
Goals of the lab:
To give students some familiarity with the Binomial distribution.
Background:
The binomial distribution is a fairly common discrete random variable. There are four conditions that must hold for
a RV to be a binomial random variable.
• There must be a fixed number of trials, n.
• Each trial can result in one and only one of two possible outcomes, labeled “success” and “failure”.
• The probability of success in a single trial, p, is the same for each trial.
• The trials are independent, so that the probability of success is unaffected by the result of a previous trial.
The probability distribution function is given below*. The textbook we are using this semester has a table that
requires some elaboration. Table G in our textbook "Statistics The Craft of Data Collection, Description, and
Inference (by Monrad, et.al.) is a cumulative binomial table. This means that the values in the table are P(Y£ r) for
some r (and, of course, n and p). Consequently if we want to calculate other binomial probabilities, then we must do
some calculations. The table below illustrates those rules.
Relevant Formulae:
* P(X
Ê nˆ x
n-x
= x) = Á p (1 - p)
for x=0,1,2,…,n
Ë x¯
Rules for Table G in textbook:
Probability we want
P(Y £ r)
P(Y < r)
P(Y ≥ r)
P(Y > r)
*P(Y=r)
Calculation we
need to perform
Example
P(Y£ r)
P(Y£(r-1))
1-P(Y£(r-1))
1-P(Y£r)
P(Y£r)-P(Y£(r-1))
P(Y£4)
P(Y<4) = P(Y£3)
P(Y≥4) = 1-P(Y£3)
P(Y>4) = 1-P(Y£4)
P(Y=4) = P(Y£4)-P(Y£3)
Examples:
There are four conditions that must hold for a RV to be a binomial random variable.
• There must be a fixed number of trials, n.
• Each trial can result in one and only one of two possible outcomes, labeled “success” and “failure”.
• The probability of success in a single trial, p, is the same for each trial.
• The trials are independent, so that the probability of success is unaffected by the result of a previous trial.
Suppose that n = 20, p = 0.35, and X = the number of successes.
P(X=6) =
ÊÁ nˆ x
Ê 20ˆ
p (1 - p)n- x = Á
(0.35)6 (1 - 0.35)20-6
Ë x¯
Ë 6¯
Ê 20ˆ
0.3560.6514
Ë 6¯
=Á
= (38760)*(0.001838)*(0.002403) = 0.1712
We can also use Table G in textbook to calculate this:
P(X=6) = P(X£6) – P(X£5) = 0.4166 – 0.2454 = 0.1712
Other examples that use the table of cumulative binomial probabilities:
P(X>3) = P(X≥4) = 1- P(X£ 3) = 1 – 0.0444 = 0.9556
P(X<10) = P(X£9) = 0.8782
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