## Math 105, Topics in Mathematics |
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## Lesson 5 :Binomial Random VariableSatya Mandal
Due Date:
See the Lecture Notes Site.
## 5.1 Binomial Trials and Number of SuccessPerhaps, the simplest statistical experiment is a Bernoulli trial. Examples of Bernoulli trials begin with tossing a coin, because it has exactly two possible outcomes - Head or Tail. Any statistical experiment that has exactly two outcomes (to be called "success" and "failure") is called a Bernoulli Trial. The experiment of tossing a coin is synonymous to any real life Bernoulli TRIAL. Other examples include testing an item for defectiveness, asking a voter whether he/she will vote for a particular candidate, a pregnancy test, testing a patient for a particular ailment and others. For an unbiased coin, the probability of "success" and "failure" are equal. In real life Bernoulli trials, such as the above examples, probability of success and failure would not be equal. Real life Bernoulli trials are synonymous to tossing a loaded coin. Such a trial is called a Bernoulli trial, which has the sample space S = {s, f} where s denotes the outcome "success" and f denotes the outcome "failure." Given a Bernoulli trial, we can define a random variable Y as follows:
Let us also have the notations
So,
This random variable Y is called the Bernoulli(p) random variable and the corresponding statistical experiment is called Bernoulli(p)-trial. Bernoulli trials are too simple to be of any serious interest. Things become interesting when we perform a series identical and independent Bernoulli(p) trials. Definition. Let n be a positive integer and 0 ≤ p ≤ 1. A Binomial(n,p)-experiment (or B(n,p)-experiment) is defined to be a combination of n "identical and independent" Bernoulli(p) trials. More formally, B(n,p) experiment is characterized as follows:- A binomial experiment consists of n identical and independent Bernoulli trials.
- The probability of success in each trial remains fixed and is equal to p.
X = total number of successes in these n trials. Then X is called a binomial (n,p)-random (or B(n,p)-random) variable. Following facts about a B(n,p)-random variable X will be useful subsequently: - X can assume values 0,1,…,n. The probability distribution
is given by
p(r) = P(X = r) = P(r success) = _{n}C_{r}p^{r}(1-p)^{n-r}
where r runs through 0,1,2,…,n.
- The mean of X is
μ = np. Note, themean =μ = (number of trial)*(probability of success in each trial). - The standard deviation of X is
σ =√np(1-p) Note, theStandard Deviation =σ = √ (number of trials)*(probability of success in each trial)*(probability of failure in each trial). - Like most random variables in nature,
Binomial(n,p)-random variables also behave approximately like Normal
random variables, when n is large and p is not too close to 0 or 1.
The following animation demonstrates the same.
Animation 5.1.1 - The graph maintains a bell-shape.
- The graph also peaks at (or near) the mean μ = np.
## Probability Computations For conceptual reasons, in classrooms we exhibit the above formulas,
to compute probability for binomial random variables. However, in real life
statistical situations this is not worthwhile. In statistics, approximation
and usage of computational tools is the name of the game.
## 5.2 Normal Approximation to BinomialBefore computers and others computational tools (like TI-84) invaded the work places and classrooms, Normal Distribution was used almost universally to approximate probability for any random variable. Even now, for huge probability computations smaller computers (say TI-84) may not be adequate. However, Normal distribution provides much more conceptual ideas and is more accurate. The Animation 5.1.1 demonstrates and justifies the same. For this course, Normal probability distribution will be used as an alternative method to approximate binomial probability, as in the next theorem. Theorem. Suppose X is a B(n,p) random variable. Assume n is large and p is not very close to 0 or 1. Then X behaves, approximately, like a N(μ, σ) random variable, with mean μ = np and standard deviation σ = √ np(1-p). For r=0,1,…,n probability is approximated as follows: P(X = r) = P(r-0.5 < X < r + .5)
≈ P(L < Z
< R) More generally, for r,s=0,1,…,n P(r ≤ X ≤ s) = P(r-0.5 < X < s +
.5) ≈ P(L < Z < R) Remark. This adjustment by .5 on two sides is called continuity correction. Recall P(Y=r) =0 for any continuous random variable Y. Because of this, we could not have treated, naively, X as Normal, without continuity "correction". On the other hand, when n is very large, it would not make any noticeable difference. Problems on 5.2:TI-84 and Normal Approximation Exercise 5.2.1. A Lawrence bank knows that
35 percent of its customers will visit the drive-through window. If
400 customers visit the bank, what is the approximate probability that
more than 120 will visit the drive-through window? Solution by TI-84: Alternative Normal Approximation: Exercise 5.2.2. It is known that the probability that a household owns a food processor is 0.1. If 190 households are interviewed, find the approximate probability that - more than 26 households own a food processor;
- less than 30 households own a food processor.
Solution by TI-84: -
Now "X is more than 26" means "X > 26", which is "X ≥27".
P(27 ≤ X) = 1 - P(X ≤ 26) = 1 -binomialcdf(190, .1, 26) = 1 - .9602 = .0398 -
Again, "X is less than 30" means "X ≤ 29".
P(X ≤ 29)= binomialcdf(190, .1, 29) = .9918
Alternative Normal Approximation: -
Now "X is more than 26" means "X > 26", which is "X ≥27".
P(27 ≤ X) = P(26.5 ≤ X) [This is continuity correction.] = P([26.5 - μ]/σ < [X - μ]/σ ) ≈ P([26.5 - μ]/σ < Z ) = P([26.5 - 19]/4.1352 < Z ) = P (1.8137, < Z ) = normalcdf(1.8137, 5)= .0349 -
Again, "X is less than 30" means "X < 30", which is "X ≤ 29".
P(X ≤ 29)= P(X ≤ 29.5) [This is continuity correction.] = P([X - μ]/σ < [29.5 - μ]/σ) ≈ P(Z < [29.5 - μ]/σ) = P(Z < [29.5 -19 ]/4.1352) = P ( Z < 2.5392) = normalcdf(-5, 2.5392) = .9944
Exercise 5.2.3. The campaign committee of
a candidate claims that sixty percent of the voters are in favor of
the candidate. You interview 150 voters. Assuming that the campaign
committe's claim is accurate, what is the approximate probability that
less than 77 will favor the candidate? Solution by TI-84: Alternative Normal Approximation: Exercise 5.2.4. A technique is used to fertilize eggs in a fertility clinic laboratory. It is known that the probability that an egg will be fertilized by this technique is 0.1. If 500 eggs are treated, what is the probability that at least 60 eggs will be fertilized? Solution by TI-84: Alternative Normal Approximation: Exercise 5.2.5. The probability that a computer chip produced in a factory is defective is is .2. If you have a sample of 60 chips, what is the probability that the number of defective chips will be less than 20? Solution by TI-84: Alternative Normal Approximation: Exercise 5.2.6. The probability that a light bulb produced by a machine is defective is p = 0.2. Suppose a quality control inspector takes a sample of 120 bulbs. What is the probability that more than 30 bulbs will be defective? Solution by TI-84: Alternative Normal Approximation: Exercise 5.2.7. Suppose the probability that a student has access to the Internet is p = 0.8. Suppose you interview 160 students. What is the probability that less than 120 students will have access to the Internet? Exercise 5.2.8. Suppose that the probability that a person favors medical use of marijuana is p = 0.6. If 780 individuals are interviewed, what is the probability that less than 450 will be in favor? Exercise 5.2.9. Suppose that the probability that a middle-income family invests in the stock market is p = 0.8. If we interview 880 middle-income families, what is the probability that more than 700 have invested in the stock market? Exercise 5.2.10. Suppose that an insurance company knows from experience that the probability that a life-insurance policyholder will survive another 10 years is p = 0.9. The company has 2280 policyholders. What is the probability that more than 2025 will survive another 10 years. |