Binomial and Related Distributions
學生 : 黃柏舜
學號 : 102581010
授課老師 : 蔡章仁
Binomial and Related Distributions
In this section of the website, we explore the binomial distribution and, in
particular, how to do hypothesis testing using the binomial distribution. We also
explain the relationship between the binomial and normal distributions, as well
as some related distributions, namely the proportion, negative binomial,
geometric, hypergeometric, beta, multinomial and Poisson distributions.
Binomial Distribution
Definition 1: Suppose an experiment has the following characteristics:
 the experiment consists of n independent trials, each with two mutually
exclusive outcomes (success and failure)
 for each trial the probability of success is p (and so the probability of failure is
1 – p)
Each such trial is called a Bernoulli trial. Let x be the discrete random
variable whose value is the number of successes in n trials. Then the probability
distribution function for x is called the binomial distribution, B(n, p), and is
defined as follows:
where C(n, x) =
Functions.
and n! = n(n–1)(n–2)⋯3∙2∙1 as described in Combinatorial
C(n, x) can be calculated by using the Excel function COMBIN(n,x).
Binomial Distribution
Observation: Figure 1 shows a graph of the probability density function for B(10, .25).
Figure 1 Binomial distribution
That the graph looks a lot like the normal distribution is not a coincidence, as we will
see shortly.
Property 1:
Binomial Distribution
Excel Function: Excel provides the following functions regarding the binomial distribution:
BINOMDIST(x, n, p, cum) where n = the number of trials, p = the probability of success
for each trial and cum takes the values TRUE or FALSE.
BINOMDIST(x, n, p, FALSE) = probability density function f(x) value at x for the
binomial distribution B(n, p), i.e. the probability that there are x successes in n trials
where the probability of success on any trial is p.
BINOMDIST(x, n, p, TRUE) = cumulative probability distribution F(x) value at x for
the binomial distribution B(n, p), i.e. the probability that there are at
most x successes in n trials where the probability of success on any trial is p.
Binomial Distribution
Example 1: What is the probability that if you throw a die 10 times it will come up six 4
times?
We can model this problem using the binomial distribution B(10, 1/6) as follows
Alternatively the problem can be solved using the Excel function:
BINOMDIST(4, 10, 1/6, FALSE) = 0.054266
Hypothesis Testing for Binomial Distribution
Example 1: Suppose you have a die and you suspect that it is biased towards the number
3, and so run an experiment in which you throw the die 10 times and count that the die
comes up 4 times with the number 3. Determine whether the die is biased.
The population random variable x = the number of times 3 occurs in 10 trials has a
binomial distribution B(10, π) where π is the population parameter corresponding to the
probability of success on any trial. We define the following null hypothesis.
H0: π ≤ 1/6; i.e. the die is not biased towards the number 3
H1: π > 1/6
Now setting α = 0.05
P(x ≤ 4) = BINOMDIST(4, 10, 1/6, TRUE)
= 0.984538 > 0.95 = 1 – α.
And so we reject the null hypothesis with 95% level of confidence.
Hypothesis Testing for Binomial Distribution
Example 2: We suspect that a coin is biased towards heads. When we toss the coin 9
times, how many heads need to come up before we are confident that the coin is biased
towards heads?
We use the following null hypothesis:
H0: π ≤ .5
H1: π > .5
Using a confidence factor of 95% (i.e. α = .05),
we calculate
CRITBINOM(n, p, 1–α) = CRITBINOM(9, .5, .95) = 7
And so 7 is the critical value. If 7 or more heads come up then we are 95% confident that
the coin is biased towards heads, and so can reject the null hypothesis.
Note that BINOMDIST(6, 9, .5, TRUE) = .9102 < .95, while BINOMDIST(7, 9, .5, TRUE)
= .9804 ≥ .95.
Relationship between Binomial and Normal Distributions
Theorem 1: If x is a random variable with distribution B(n, p), then for sufficiently largen,
the distribution of the variable
where
Corollary 1: Provided n is large enough, N(μ,σ) is a good approximation for B(n, p) where
μ = np and σ2 = np (1 – p).
Observation: The normal distribution is a good approximation for the binomial
distribution when np ≥ 5 and n(1 – p) ≥ 5. Another way to look at this, is that the normal
distribution is a good approximation for the binomial distribution when n > 10 and .4
< p < .6, or n > 30 and .1 < p < .9.
Example 1: What is the normal distribution approximation for the binomial distribution
where n = 20 and p = .25 (i.e. the binomial distribution displayed in Figure 1 of Binomial
Distribution)?
As in Corollary 1, define the following parameters:
Since np = 5 ≥ 5 and n(1 – p) = 15 ≥ 5, based on Corollary 1 we can conclude that B(20,.25)
~ N(5,1.94).
We now show the graph of both pdf’s to see visibly how close these distributions are:
Figure 1 – Binomial vs. normal distribution
Proportion Distribution
Definition 1: If x is a random variable with binomial distribution B(n, p) then the random
variable y = x/n is said to have the proportion distribution.
Property 1: Where y has a proportion distribution as defined above
Proof: By Property 1b of Expectation and Property 1a of Binomial Distribution
By Property 3b of Expectation
Theorem 1: Provided n is large enough – generally when np ≥ 5 and n(1–p) ≥ 5 –
then N(μy,σy) is a good approximation for the proportion distribution for y with
Hypothesis Testing – one sample
From the theorem, we know that when sufficiently large samples of size n are taken, the
distribution of sample proportions is approximately normal, distributed around the true
population proportion π, with standard deviation (i.e. the standard error)
We can use this fact to do hypothesis testing as was done for the normal distribution. In
addition when a two-tailed test is performed a confidence interval can be calculated
where
where we use the sample mean p as an estimate for the population mean when
calculating the standard error. This introduces additional error, which is acceptable for
large values of n.
Example 1: A company believes that 50% of their customers are women. A sample of 600
customers is chosen and 325 of them are women. Is this significantly different from their
belief?
Method 1: Using the binomial distribution, we reject the null hypothesis since:
Method 2: By Theorem 1 we can also use the normal distribution
The observed mean is 325/600 = 0.541667
And so we reach the same conclusion, namely to reject the null hypothesis.
Example 2: A survey of 1,100 voters showed that 53% are in favor of the new tax reform.
Can we conclude that the majority of voters (from the population) are in favor?
NORMDIST(.53, .5, 0.01505, TRUE) = .976889 > .95, and so we can reject the null
hypothesis and conclude with 95% confidence that the population will vote in favor of
the tax reform.
We determine the 95% confidence interval as follows:
zcrit = NORMSINV(1 – α/2) = NORMSINV(0.975) = 1.96
And so the 95% confidence interval is
And so we conclude with 95% confidence that between 50.1% and 55.9% of the
population will be in favor. If however we are looking for a 99% confidence interval then
zcrit = NORMSINV(1 – α/2) = NORMSINV(0.995) = 2.58
And so the 99% confidence interval is
This means that with 99% confidence, between 49.1% and 56.9% of the population will
be in favor.
Hypothesis Testing – two samples
Theorem 2: Let x1 be a proportional distribution with mean π1 and number of trials
n1 and let x2 be a proportional distribution with mean π2 and number of trials n2. When
the number of trials n1 and n2 are sufficiently large, usually when ni πi ≥ 5 and ni(1–πi) ≥ 5,
the difference between the sample proportions p1 – p2 will be approximately normal with
mean π1 – π2 and standard deviation
Proof: Based on Theorem 2 of the Binomial Distribution, xi has approximately the
distribution
Since x1 and x2 are independently distributed, by the linear transfer property of the normal
distribution, x1 – x2 has distribution
Example 4: A company which manufactures long-lasting light bulbs sells halogen and
compact florescent bulbs. They ran an experiment in which they ran 100 halogen and 100
florescent bulbs continuously for 250 days. After 250 days they found that half of the
halogen bulbs were still working while 60% of the florescent bulbs were still operating. Is
there a significant difference between the two types of bulbs?
Let x1 = the percentage of halogen bulbs that are functional after 200 days and x2 = the
percentage of florescent bulbs that are functional after 250 days. The presumption is that
the distributions for each of these are proportional. We now test the following null
hypothesis:
H0: π1 = π2
Assuming the null hypothesis is true, then based on the null hypothesis by Theorem 2, x1 –
x2 will be approximately normal with mean π1 – π2 = 0 and standard deviation
where the common value of the mean is denoted π and both samples are of size n. Since
the value for π is unknown, we estimate its value from the sample, namely, 50 + 60 = 110
successes out of 200, i.e. π ≈ 0.55, Thus, the mean of x1 – x2 is 0 (based on the null
hypothesis) and the standard deviation is approximately
. The observed value of
x1 – x2 is .60 – .50 =.10, and so we have (two-tail test):
p-value = NORMDIST(.1, 0, .0497, TRUE) = .978 > .975 = 1 – α/2
Thus, we reject the null hypothesis, and conclude there is a significant difference
between the two types of bulbs.
We reach the same conclusion since critical value of x1 – x2 = NORMINV(.975,0,.0497)
= .0975 < .1 = observed value of x1 – x2.
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