Ch 4.3 Binomial Distribution

Requirements for Binomial Distribution:

X can be modeled by binomial distribution if it satisfies four requirements:

1. The procedure has a fixed number of trials. (n)

2. The trials must be independent.

3. Each trial has exactly two outcomes, success and failure, where x = number of success in n trials.

4. The probability of a success remains the same in all trials. P(success in one trial ) = p.

           P(failure in one trial ) = 1 – p = q

           P(X) = x number of success in n trials.

Note: for sampling, use 5% guideline for independent.

Ex1: Determine if the following X is binomial or not

a. X = number of adults out of 5 who use iPhone.

b. X = number of times a student raises his/her hand in a class.

c. X = number of one after tossing a die 7 times.

d. X = number of tosses until the “one” shows up.

e. X = the way student commute to school.

 

a and c are binomial. a = B(5, p),  c = B(7, 1/6)

b,d  does not have a fixed number of trials.

e : X is not a count of success.

 

Find P(X) or P(range of X) when X is binomial:

n = number of trials, p= P(success in one trial)

q = P(failure in one trial) = 1 - p, X = number of success.

Method 1: use formula:

\( P(x) = \frac{n!}{x!(n-x)!} p^x q^{n-x} \) 

Method 2: Use Statdisk /Analysis/ Probability Distribution/Binomial distribution

Enter n, p, x.. output in sample editor under P(x),  P(x or fewer) or P(x or greater).

Optional: use OnlineStatbook binomial calculator:

http://onlinestatbook.com/2/calculators/binomial_dist.html 

input n and p (to N and ∏), select above, below or between.

 

Parameters of binomial distribution:

mean μ = np

variance: \( σ² = npq \)

standard deviation \( σ = \sqrt{npq} \)

 

Range rule of thumb:

Values not significant: Between (μ - 2σ ) and (μ + 2σ )

 

Find parameters of binomial distribution

Use Statdisk /Analysis/ Probability Distribution/ Binomial distribution, enter n, p, x, evaluate.

Mean, standard deviation and variances are under the sample editor.

 

Ex1. In a college, 35% of all students are full-time students. If 11 students are randomly chosen.

a) Can probability of X = number of full time students out of 11 be modeled by binomial distribution?

   Ans:  yes, since 11 students is less than 5% of all students, P(one student is full time) = 0.35 = constant,  11 is a constant number of trials.

           there are two outcomes for each student, full-time or not full-time.

 

b) What is the probability that there are 4 full-time students out of 11?

Use statdisk/analysis/probability distribution/ binomial distribution n = 11, p = 0.35, x = 4, evaluate.  Use P(x)

P(x) = 0.2428, P(4 out of 11 are full-time) = 0.2428. 

 

c) What is the probability that there are less than 5 full-time students?

Use statdisk/analysis/probability distribution/ binomial distribution n = 11, p = 0.35, x = 4. use P(x or fewer)

P( x or fewer) = 0.6683. The chance of less than 5 full-time students out of 11 is 0.6683.

 

d) What is the probability of there are more than 3 full-time students?

P( more than 3) = P( 4 or more)

Use Statdisk/analysis/probability distribution/ binomial distribution n= 11, p = 0.35, x = 4 use P( x or greater)

P( x or greater) = 0.5744

 

Ex2: A bookstore manager estimates that 9.5% of all customers coming in the store will buy a book or magazine.  If 24 customers visit the store on a certain business hour,

a) Can x = number of customers out of 24 who buy a book or magazine be modeled by binomial distribution?

Yes X can be modeled by binomial distribution because there are 2 outcomes, buy book or magazine or not "buy book or magazine". 24 customers should be less than 5% of the population of all customers, so sample are independent. P(one customer buy) = 0.095 is constant.

So we can use binomial distribution n = 24, p =0.095, 

b) Find the probability that exactly 3 customers will buy a book or magazine.

Use Statdisk/analysis/probability distribution/ binomial distribution n = 24, p =0.095, x = 3, evaluate, use P( x)

P(x) = 0.2133

c)  Find the probability that at least  5 customers will buy a book or magazine.

P( at least 5) = P( 5 or more) 

Use Statdisk/analysis/probability distribution/ binomial distribution n = 24, p =0.095,  x = 5, evaluate, use P(x or greater)

P(x or greater) = 0.0714

d) Find the probability that at most 2 customers will buy a book or magazine.

P( at most 2) = (2 or fewer)

Use Statdisk/analysis/probability distribution/ binomial distribution n = 24, p =0.095,  x = 2, evaluate, use P(x or fewer)

P( x or fewer) = 0.5977

e) Find the non-significant range of customer who will buy a book or magazine out of 24 customers.

Find mean and standard deviation from statdisk/analysis/probability distribution/ binomial distribution, n = 24, p = 0.095, x = 0, evaluate

Evaluate, look at the bottom of the table.

 Mean = 2.28,  sd = 1.44, 

  Non-significant range = 2.28 – 2(1.44) = -0.60  to   2.28 + 2(1.44) = 5.16.  

  X values from -0.6 to 5.2 are non-significance.

 

Ex3. A small airline has a policy of booking as many as 60 persons on an airplane that can seat only 53. (Past studies have revealed that only 78% of the booked passengers actually arrive for the flight.)

a) Find the probability that if the airline books 60 persons, not enough seats will be available.

    Use binomial distribution n= 60, p= 0.78, P(not enough seats) = P( 54 or more)

   Use Statdisk/analysis/probability distribution/ binomial distribution n = 60, p =0.78,  x = 54, evaluate, use P(x or more)

    P( x or more) = 0.013.

b) Find the non-significant range of passengers who will arrive out of 60 passengers.

    Look at the bottom of the statdisk table, mean = 46.80, sd =3.21

    non-significant range is from 46.8 - 2(3.21) to 46.8 + 2(3.21). From 40.38 to 53.22 or 40.4 to 53.2