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PROBABILITY MODELS
TABLE OF CONTENTS
- CHAPTER - Latest Updated 2024/ Comprehensive Exam Study Guide
- 1 The Binomial Probability Distribution Page
- Conditions for a Binomial Experiment, Bernoulli Trials
- Mean and Standard Deviation of a Binomial Random Variable
- Using Excel and Table 1 to Calculate Binomial Probabilities
- 2 The Normal Probability Model
- The Standard Normal Distribution
- Standardizing Normal Distributions
- Mensa Example
- Tele-Evangelist Example
- 3 Approximating a Binomial Model with the Normal Distribution
- Advertising Campaign and Dear Abby Examples
- ESP Example
- 4 Fundamental Ideas From The Fourth Session
- Six Probability Model Exercises
- Answer Sketches for Six Probability Model Exercises
- 5 Table One: The Binomial Distribution
- 6 Table Two: The Normal Distribution
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The Binomial Probability Distribution
The simplest data-gathering process is counting the number of times a certain event occurs. We can reduce almost an endless variety of situations to this simple counting process.
Consumer preference and opinion polls (i.e., sample surveys) are conducted frequently in business. Some recent examples include polls to determine the number of voters in favor of legalizing casino gambling in a particular state, and the number of Americans in favor of government-controlled health care.
The Binomial distribution is a discrete probability distribution that is extremely useful for describing many phenomena. The random variable of interest that follows the Binomial distribution is the number of successes obtained in a sample of n observations. The Binomial distribution can be applied to numerous applications, such as:
What is the probability that in a sample of 20 tires of the same type none will be defective if 10% of all such tires produced at a particular plant are defective? What is the probability that red will come up 15 or more times in 20 spins of the American roulette wheel (38 spaces)? What is the probability that a student can pass (that is, get at least 60% correct on) a 30-question multiple-choice exam (each question containing four choices) if the student guesses each question? What is the probability that a particular stock will show an increase in its closing price on a daily basis over the next ten (consecutive) trading sessions if stock market price changes really are random?
Conditions Required for a Binomial Experiment
- There is a set of n trials, which can be classified as either “successes” or “failures.”
- The probability of a success, , is constant over the n trials.
- The outcome for any one trial is independent of the outcome for any other trial.
These first three conditions specify Bernoulli Trials.
- The Binomial random variable X counts the number of “successes” in n trials.
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and . Note that for a random variable X following the Binomial ( n , ) distribution: m (^) X E X np , and s (^) X
We can obtain binomial probabilities using a Table (like Table 1) or Excel.
np 1 p
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Using Excel to find Binomial Probabilities
To obtain the probability Pr( Y = k ) for a Binomial random variable Y with parameters n and p using Excel, use the function =BINOMDIST( k, n, p , 0). Suppose Y is a Binomial random variable with n = 5 and p = .50. To find Pr(Y = 0), enter =BINOMDIST(0, 5, 0.5, 0) in any empty cell and press Enter to obtain the probability .03125. In this way, a complete Binomial table for n = 5 and p =. can be obtained: k 0 1 2 3 4 5 Pr(Y = k) .03125 .15625 .31250 .31250 .15625.
Sometimes, we will want to obtain a cumulative (“less than or equal to”) probability. For instance, suppose we want to find the probability that Y 4. To do this, we can either add together the individual probabilities: Pr(Y 4) = Pr(Y = 0) + Pr(Y = 1) + Pr(Y = 2) + Pr(Y =
- Pr(Y = 4) OR adjust the Excel command and use the function BINOMDIST( k, n, p , 1). For instance, in this setting, we’d use =BINOMDIST(4, 5, 0.5, 1). The “1” in the last part of the command asks for a cumulative (“less than or equal to”) probability instead of a simple (“equal to”) probability. A complete cumulative probability table for the Binomial distribution with n = 5 and p = .50 is easily generated from the information above:
k 0 1 2 3 4 5 Pr(Y k) .03125 .18750 .50000 .81250 .96875 1.
Using Table 1 to find Binomial Probabilities (see pp. 135-142)
To use the Table of Binomial Probabilities [Table 1], we first select the appropriate subtable for our n and then the correct probability . The table gives the probability distribution of X , the number of successes, for a wide variety of small Binomial experiments. Note that values of below .50 can be read at the top of the block and values of x are read on the left. For values of above .5, values of are read at the bottom, and values of x on the right.
Suppose n = 5 and = .30. Here’s the n = 5 section of Table 1, pulled from the table on page 135.
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0 .7738^ .5905^ .4437^ .3277^ .2373^ .1681^ .1160^ .0778^ .0503^ .0313^ 5 1 .2036^ .3281^ .3915^ .4096^ .3955^ .3602 .3124^ .2592^ .2059^ .1563^ 4 2 .0214^ .0729^ .1382^ .2048^ .2637^ .3087^ .3364^ .3456^ .3369^ .3125^ 3 3 .0011^ .0081^ .0244^ .0512^ .0^879 .1323 .1811^ .2304^ .2757^ .3125^ 2 4 .0000^ .0005^ .0022^ .0064^ .0146^ .0284^ .0488^ .0768^ .1128^ .1563^ 1 5 .0000^ .0000^ .0001^ .0003^ .0010^ .0024^ .0053^ .0102^ .0185^ .0313^ 0 .95 .90 .85 .80 .75 .70 .65 .60 .55 .50 k
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Example 4.1. A Simple Binomial Distribution Example
Let X be a Binomial random variable with n = 4 trials and probability of success = .70. Let x be the number of successes observed.
(a) Find Pr( x = 3). ( b ) Find Pr( x < 2). ( c ) Find Pr( x 2). ( d ) What would be the modal (most common) outcome (number of successes)?
We could, for instance, draw a tree of the n = 4 trials, as shown on the next page. From the tree, we can directly find the probability distribution of X. For instance, PX(0) = Pr( X = 0) = .0081, and PX(1) = Pr(X = 1) = 4(.0189) = .0756, etc.
In Excel , we can use statements of the form =BINOMDIST(k, 4, 0.7, 0) to find the simple (“equal to”) probability distribution for X. For instance, =BINOMDIST(0, 4, 0.7, 0) produces 0. =BINOMDIST(1, 4, 0.7, 0) produces 0. =BINOMDIST(2, 4, 0.7, 0) produces 0.2646, etc. As before, we can find cumulative probabilities using =BINOMDIST(k, 4, 0.7, 1).
To use Table 1 , we look at the n = 4 section of the table on page 135, and = .70, which is reproduced (and shaded) here: n = 4 k p = .
.10 .15 .20 .25 .30 .35 .40 .45.
0 .8145^ .6561^ .5220^ .4096^ .3164^ .2401 .1785^ .1296^ .0915^ .0625^ 4 1 .1715^ .2916^ .3685^ .4096^ .4219^ .4116^ .3845^ .3456^ .299^5 .2500^ 3 2 .0135^ .0486^ .0975^ .1536^ .2109^ .2646 .3105^ .3456^ .3675^ .3750^ 2 3 .0005^ .0036^ .0115^ .0256^ .0469^ .0756^ .1115^ .1536^ .2005^ .2500^ 1 4 .0000^ .0001^ .0005^ .0016^ .0039^ .0081 .0150^ .0256^ .0410^ .0625^ 0 .95 .90 .85 .80 .75 .70 .65 .60 .55 .50 k
Note that we must read up from the bottom (since here is .70, not .30), we have Number of successes, x 0 1 2 3 4
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(b) Pr(X < 2) = Pr(X = 0) + Pr(X = 1) = .0081 + .0756 = .0837. (c) Pr(X 2) = Pr(X = 2) + Pr(X = 3) + Pr(X = 4) = .2646 + .4116 + .2401 = .9163. (d) There are five possible outcomes: 0, 1, 2, 3 or 4 successes. The outcome with the highest probability is 3 successes: Pr(X = 3) = .4116, which is higher than Pr(X = 0) = .0081, Pr(X = 1) = .0756, Pr(X = 2) = .2646, or Pr(X = 4) = .2401.
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Trial One Trial Two Trial Three Trial Four X (# of S’s) Probability S on 4 th^ (.70) 4 (.7)^4 =. S on 3 rd^ (.70) F on 4 th^3 (.7)^3 (.3)^1 = (.30). S on 2 nd (.70) S on 4 th^ (.70) 3 (.7)^3 (.3)^1 = . F on 3 rd (.30) Success (S) on (^) F on 4 th (.30)
2 (.7)^2 (.3)^2 = . 1 st^ Trial (.70 = probability)
S on 4 th^ (.70) 3 (.7)^3 (.3)^1 = . S on 3 rd^ (.70) F on 4 th (.30)
2 (.7)^2 (.3)^2 = . F on 2 nd (.30) S on 4 th^ (.70) 2 (.7)^2 (.3)^2 = . F on 3 rd (.30) F on 4 th (.30)
1 (.7)^1 (.3)^3 = .
S on 4 th^ (.70) 3 (.7)^3 (.3)^1 = . S on 3 rd^ (.70) F on 4 th (.30)
2 (.7)^2 (.3)^2 = . S on 2 nd (.70) S on 4 th^ (.70) 2 (.7)^2 (.3)^2 = . F on 3 rd (.30) F on 4 th (.30)
1 (.7)^1 (.3)^3 = . Failure (F) on 1 st^ Trial S^ on^4 th^ (.70)^2 (.7)^2 (.3)^2 = . (.30 = S on 3 rd^ (.70) probability) F on 4 th (.30)
1 (.7)^1 (.3)^3 = . F on 2nd (.30)
Statistics Pre-Term Program – T. Love & V. Babich – Chapter 4 - Page 118
S on 4th^ (.70)
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Answering the Four Questions from the Start of the Chapter
What is the probability that in a sample of 20 tires of the same type none will be defective if 10% of all such tires produced at a particular plant are defective?
Consider each tire inspection to be a Bernoulli trial, with the event “sampled tire is defective” defined as a success. We need the probability of 0 successes in n = 20 trials, with = Pr(success) = .10 The relevant Table 1 section follows. The probability of k = 0 successes is.. n = 20 k p =. 0.
What is the probability that red will come up 15 or more times in 20 spins of the American roulette wheel (38 spaces)?
Consider each spin of the wheel to be a Bernoulli trial, with the event “ball stops in red space” defined as a success. Here, we want to know the probability of 15 or more successes in n = 20 trials. If each of the 38 spaces is equally likely, then since there are 18 red spaces, the probability of a success is p 18 9 .474. Since Table 1 does not give exact probabilities for this value of , let’s use Excel 38 19 to solve the problem.
As we have seen, the appropriate Excel statement to find the “equal to k ” probability distribution for X is of the form =BINOMDIST(k, n, , 0). In this case, we have n = 20, = 9/19, and k = 15, …, 20.
Probability Statement Excel Statement ( n = 20, = 9/19) Result Pr(X = 15) =BINOMDIST(15, 20, 9/19, 0) 0. Pr(X = 16) =BINOMDIST(16, 20, 9/19, 0) 0. Pr(X = 17) =BINOMDIST(17, 20, 9/19, 0) 0. Pr(X = 18) =BINOMDIST(18, 20, 9/19, 0) 0. Pr(X = 19) =BINOMDIST(19, 20, 9/19, 0) 7.59E- 06 Pr(X = 20) =BINOMDIST(20, 20, 9/19, 0) 3.23E- 07
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Summing these results, the probability of 15 or more “red” is approximately.
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probability that:
a. Exactly five of the swordfish pieces have mercury levels above the FDA maximum.
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b. At least ten swordfish pieces have a mercury level above the FDA maximum.
c. At most seven swordfish pieces have a mercury level above the FDA maximum.
(^1) Sincich, Exercise 5.22.
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2 ps^2 ps
2
The Normal Probability Model
We now turn our attention to continuous probability density functions, which arise due to some measuring process on a phenomenon of interest. When a mathematical expression is available to represent some underlying continuous phenomenon, the probability that various values of the random variable occur within certain ranges or intervals may be calculated. However, the exact probability of a particular value from a continuous distribution is zero. A particular continuous distribution, the Normal, or Gaussian , model is of vital importance.
- Numerous continuous phenomena seem to follow it or can be approximated to it.
- We can use it to approximate various discrete probability distributions and thereby avoid much computational drudgery.
- It provides the basis for classical statistical inference because of its relationship to the Central Limit Theorem.
Crucial theoretical properties of the Normal distribution include:
- It is bell-shaped and symmetrical in its appearance.
- Its measures of central tendency (mean, median and mode) are all identical.
- The Empirical Rule holds exactly for the Normal model.
The Normal probability density function is: (^ y ^ m^ )
(^2) (^) (^) f (^) Y ( y )
1 1 ( y m )^2 s^2 1
e
^2 s 2
The values and in the normal density function are in fact the mean and standard deviation of Y. The Normal distribution is completely determined once the parameters and are specified. A whole family of normal distributions exists, but one differs from another only in the location of its mean and in the variance ^2 of its values.
e
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The Standard Normal Distribution (Table 2)
Tables of normal curve areas (probabilities) are always given for the Standard Normal Distribution. With a little effort, we can calculate any Normal probability based on just this one table.
