Chapter 14/15 Binomial Distribution

Properties– Two possible outcomes (success and failure)– A fixed number of experiments (trials)– The probability of success, denoted by p, is

the same on every trial– The trials are independent

• Example: Suppose 75% of all drivers wear their seatbelts. Find the probability that four drivers might be belted among five cars waiting traffic light?

Chapter 14/15 Binomial Distribution

• We look for the number of k successes in

n trial. Here k is less than or equal to n. • Let X = number of successes in n trials. • P(X=x) = C(n,x)*p^x*q^{n-x}• Here p = probability of success, q = 1-p =

probability of failure.

Chapter 14/15 Binomial Distribution

• Example: Suppose 20 people come to the blood drive. What is the probability that there are 2 or 3 universal donors?

Solution: • P(X=2) = C(20,2)(.06)^{2}(.94)^{18}=

0.2246• P(X=3) = C(20,3)(.06)^{3}(.94)^{17}=

0.0860• Answer: 0.3106

Chapter 14/15 Binomial Distribution

• To compute the C(n,x) number from the TI-83 do the following

• 1. Type your number• 2. Press MATH• 3. Select Prob• 4. Press nCr• 5. Type your second number

Examples

2. A certain tennis player makes a successful first serve 70% of the time. Assume that each serve is independent of the others. If she serves 6 times, what’s the probability she gets

a) all 6 serves inb) exactly 4 serves in?c) at least four serves in? d) no more than 4 serves in?

Examples2 binomial model: p = 0.7, q = .3, n = 6

a) C(6,6)*.7^6*.3^0 = 0.118b) C(6,4)*.7^4*.3^2 = 15*.7^4*.3^2 = .324c) C(6,4)*.7^4*.3^2 = 0.324

C(6,5)*.7^5*.3^1 = 0.303C(6,6)*.7^6*.3^0 = 0.118

Answer: 0.745

d) 1 – (.303+.118) = .575

Example

A statistics test contains 5 multiple choicequestions, each of which has four choices.Suppose a student guesses the answer toeach question. Find the probabilitydistribution of X, the number of questionsthe student answers correctly.

Example

X=k 0 1 2 3 4 5

P(X=k) .2373 .3955 .2636 .0878 .0146 .0009

Concluding Remarks on

1. Binomial as an Approximation to Hypergeometric Distribution

Example: A city has 1000 residents of whom 450 are male. 200 are to be selected at random without replacement.

Clearly this is a hypergeometric distribution problem with parameters:

N = 1000, n = 450, M = 200.

Binomial Approximation of Hypergeometric Distribution

• Compute the probabilities for the following values of m:

• m = 90, 91, 92, …, 110.

Example: A city has 10 residents of whom 4 are male. 5 are to be selected at random without replacement.

This is a hypergeometric distribution problem with parameters: N = 10, n = 4, M = 5.

Compare the hypergeometirc and binomial distributions.

Concluding Remarks

2. How much variation is typical? Example:

Blood is drawn and a blood count isperformed by measuring two things: # of white blood cells proportion of the different types of white blood

cells. These measurements are compared to typicalmeasurements to determine whether there iscause for concern.

Concluding Remarks

Example (neutrophil)

Suppose that the typical proportion of\neutrophils is 0.6. In a blood count it wasfound that 45 out of 100 white blood cellswere neutrophils. Is there enough evidenceto be concerned? In other words the number45 typical or not?

Concluding Remarks

We can model this by the binomialdistribution.

Let X = # of neutrophils in 100 white blood cells drawn from a person whose proportion of neutrophils is normal (i.e. 0.6).

We would like to study the probabiltydistribution

Concluding Remarks

Question: Is it surprising to find a blood count different from 60?

Let us find the likelihood of getting 60neutrophils in a 100 blood count (if weassume that the proportion of neutrophils ina normal person is 0.6)

Concluding Remarks

• This can be modeled by the binomila distribution.

P( X = 60) = C(100,60)(.6^60)*(.4^(40) =0.081

• Hence it is not surprising to find number neutrophils to be other than 60

• How surprising is it to find 45 neutrophils?

Concluding Remarks

• Better question will be: How likel is it to find the number of neutrophils to be less than 45?

• We compute P(X <= 45) = .0017.

• Hence it is very unlikely that this happens under normal situations. The differnece between 60 and 45 is probably significant!

• Cause to be concerned is probably justified!

Concluding Remarks

• When we compute P(X < = m), we are finding what is called “a cumulative distribution function”.

• Example: Let X be binomial with parameters n = 5 and p = 0.4. Find the cumulative distribution function of X.

• Note: m = 0, 1, 2, 3, 4, 5.

Concluding Remarks

m 0 1 2 3 4 5

P(X=m) 0.07776 0.2592 0.3456 0.2304 0.0768 0.01024

P(X<=m) 0.07776 0.33696 0.68256 0.91296 0.98976 1

Concluding Remarks

3. Is selection random? (Historical case: Discrimination against Mexican-Americans-Read the course pack (page 102)

ProblemSuppose in a county (Hidalgo County, TX)79.1% of the population is of Hispanic origin.From 1962-1972, 870 were summoned toserve on a grand jury. Of these 339 had

Spanish surnames. How likely is this if the juryselection was random?

Concluding Remarks

• Let X = # of Hispanics who served on a grand jury.

• Compute P(X < = 339)

n = 870, p = .791, m = 0, 1, 2, 3, 4, …, 339.

P(X < = 339) = 4.18*10^(-148) ~ 0. This will be the probability if the jury selection was random. Exremely unlikely!