Section 5.2

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Section 5.2

Binomial Distribution

With some useful additional content by D.R.S., University of Cordele.

Recognize Binomial Situations

Only two possible outcomes in each trial.– Probability for one of the outcomes.– Probability for the other outcome.

Some definite number of trials, .– They’re independent trials. don’t change.

We’re interested in , probability of a certain count of how many times event happens in those trials.

A special kind of probability distribution

It’s the familiar probability distributionBut only two rows for the two outcomes.Note that

– Because probabilities must always sum to 1.00000– And this leads to .

Outcomes Probabilities

One of the events

The other event

Total Exactly 1






Probability for a Binomial Distribution

Probability for a Binomial Distribution For a binomial random variable X, the probability of obtaining x successes in n independent trials is given by

where x is the number of successes, n is the number of trials, and p is the probability of getting a success on any trial.

1 n xxn xP X x C p p

Practice with the Formula

Experiment: Roll two diceEvent of interest: “I rolled a 7 or an 11”Binomial! Either-Or. Two outcomes. Win or Lose.Probability of success: (from Probability of failure: Number of trials

Practice with the Formula

Find P(2) successes in the seven/eleven game

How about 3 successes?

Compute them all

Summary of the 7-11 experiment

X successes P(X successes)

0 times (no sevens or elevens)

1 time

2 times

3 times

4 times

5 times (all sevens and elevens)

Total (must equal 1.000000 !!)

Sometimes you add probabilities

Probability of at least three wins in five trials– P(X≥3) = P(X=3) + P(X=4) + P(X=5) add them up!

Probability of more than three wins– P(X>3) = P(X=4) + P(X=5)

Probability of at most three wins – P(X≤3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

Probability of fewer than three wins– P(X<3) = P(X=0) + P(X=1) + P(X=2)

Use the Complement to save time

Example: “Probability of at least 3 wins”P(X≥3) =

P(X=3) + P(X=4) + … + P(X=49) + P(X=50)This means 48 calculations and sum results.EASIER: The complement is “fewer than 3”Take 1 – [ P(X=0) + P(X=1) + P(X=2) ]

TI-84 Computations

binompdf(n, p, X) = probability of X successes in n trialsRecompute the table and make sure we get the same results as the by-hand calculations.The “pdf” in “binompdf” stands for “probability distribution function”

Excel: =BINOM.DIST(X, n, p, FALSE)Note different order from TI-84 binompdf(n, p, X)

Excel =BINOM.DIST(X, n, p, FALSE)

Note different order from TI-84 binompdf(n, p, X)

Mean, Variance, and Standard Deviation

We had formulas and methods for probability distributions in general.The special case of the Binomial Probability Distribution has special shortcut formulas

– Mean = – Variance = – Standard deviation =


Compute these for the seven-eleven experiment with n = 5 trials




– Mean = and Standard deviation =

“Expected value” – in 100 tosses of two dice, how many seven-elevens are expected?

– Remember, the mean of a probability distribution is also called the “expected value”

Standard Deviation

What happens to the standard deviation in the seven-eleven experiment as the number of trials, n, increases?

trials Standard deviation

5

50

100

TI-84 Computations

binompdf(n, p, x) = P(X=x) successes in n trialsObserve the difference between binompdf() and binomcdf()

binomcdf(n, p, x) = P(X=0) + P(X=1) + … + P(X=x) successes in n trialsbinomcdf(n, p, x) does lots of little binompdf() for you for x = 0, x = 1, etc. up to the x you told it, and it adds up the resultsThe “pdf” stands for “probability distribution function”The “cdf” stands for “cumulative distribution function”

Advanced TI-84 Exercise

Y1=binompdf(20,8/36,X)seq(X,X,0,20) STO> L1

seq(Y1(X),X,0,20) STO> L2

STAT PLOT for these two lists, histogramWINDOW Xmin=0, Xmax=20, Ymin=-0.1,Ymax=0.6ZOOM 9:ZoomStat

Try and verify binomcdf(n,p,x)

X successes P(X successes)Using binompdf

P(0 thru X) successesUsing binomcdf

0 times (no sevens or elevens)1 time

2 times

3 times

4 times

5 times (all sevens and elevens)Total (must equal 1.000000 !!)

Excel: =BINOM.DIST(X,n,p,TRUE) for cumulative

Add 0.28463+0.40661+0.23235 and verify it = 0.92359

binomcdf() and complements

Sevens or elevens, n = 50 trials againP(no more than 10 successes)

– binomcdf(50, 8/36, 10)P(fewer than 10 successes)

– binomcdf(50, 8/36, 9)P(more than 10 successes) – use complement!

– 1 minus binomcdf(50, 8/36, 10)P(at least 10 successes) – use complement!

– 1 minus binomcdf(50, 8/36, 9)






Example 5.5: Calculating a Binomial Probability Using the Formula

What is the probability of getting exactly six heads in ten coin tosses? SolutionWe showed earlier that coin tosses meet the criteria of the binomial distribution. For this problem, let X = the number of heads obtained out of ten coin tosses. There are ten coin tosses, so n = 10. We will say that a success is getting a head. We want the probability of exactly six successes, so x = 6. The probability of flipping a head in one coin toss is 0.5, which means that p = 0.5.






Example 5.5: Calculating a Binomial Probability Using the Formula (cont.)

Substituting these values into the binomial probability formula gives us the following.

6 10 610 6

6 4

6 0.5 1 0.510!

0.5 0.56!4!210 0.015625 0.0625

05

1

0.2 1

n xxn xP X x

P

C p p

X C






Example 5.5: Calculating a Binomial Probability Using the Formula (cont.)

Thus, the probability of getting exactly six heads in ten coin tosses is approximately 0.2051.






Example 5.6: Calculating a Binomial Probability Using the Formula or a TI-83/84 Plus Calculator

A quality control expert at a large factory estimates that 10% of all batteries produced are defective. If he takes a random sample of fifteen batteries, what is the probability that exactly two are defective?SolutionFirst, let’s verify that this process meets the criteria of a binomial distribution. Since the batteries are randomly sampled and, presumably, more batteries continue to be produced by the factory while the sampling takes place, we can consider the selection of the batteries to be identical, independent trials.






Example 5.6: Calculating a Binomial Probability Using the Formula or a TI-83/84 Plus Calculator (cont.)

Since we are testing fifteen batteries, the number of trials is n = 15. For each trial, there are two possible outcomes: either the battery is defective or it is not. We will consider a defective battery to be a success and 10% of all batteries produced are defective, so the probability of getting an individual success is p = 0.1. Let X = the number of defective batteries found in a sample of 15 batteries. We are looking for the probability that exactly two are defective, so we want the binomial probability, P(X = 2).







Using the binomial probability formula for our solution would require us to calculate the following expression.

However, the TI-83/84 Plus calculator can calculate P(X = x) directly using the following procedure.

2 1315 22 0.1 00

..2669

1

9

n xxn x

P X

P X x C p p

C







• Press and then to access the DISTR menu.

• Choose option A:binompdf(. • Enter n, p, and x in the parentheses as: binompdf (n, p, x). Thus, using a TI-83/84 Plus, we would calculate the probability as shown below and in the screenshot in the margin.







Therefore, the probability that exactly two out of the fifteen batteries are defective is approximately 0.2669.

0.2

, ,

2669

P X x

X

n x

P

p

bi nompdf ( 15, 0. 1,2)bi nompdf ( )






Example 5.7: Calculating Binomial Probabilities Using the Formula or a TI-83/84 Plus Calculator

A quality control expert at a large factory estimates that 10% of all the batteries produced at the factory are defective. If he takes a random sample of fifteen batteries, what is the probability that no more than two are defective? SolutionThis scenario is the same as in the previous example; therefore, we know that we have a binomial distribution with n = 15 and p = 0.1. This time we want the probability that no more than two are defective, which is

2 .P x






Example 5.7: Calculating Binomial Probabilities Using the Formula or a TI-83/84 Plus Calculator (cont.)

Thus we are looking for the probability that X = 0, or X = 1, or X = 2. We can find by adding these three individual probabilities.

Using the binomial probability formula for our solution would require us to calculate the following expression.

2 .P x

2 0 1 2P X P X P X P X







A TI-83/84 Plus calculator can calculate P(X = x) directly as seen in the previous example. Thus, using a TI-83/84 Plus, we would calculate the probability as shown below and in the screenshot in the margin.

0 15 1 14

15 0 15 1

2 1315 2

0.1 0.9 0.1 0.9

0.1 0

2 0 1 2

.9

0.8159

P X P X P X P X

C C

C







However, the TI-83/84 Plus calculator allows us to use an even more efficient method for this particular problem, as it will also directly calculate a cumulative probability,

0.8159

= ( )+ ( )+

2 0 2

( )

1P X P X P X P Xbi nompdf 15,0. 1,0 bi nompdf 15,0. 1,1bi nompdf 15,0. 1,2

.P X x







• Press and then to access the DISTR menu.

• Choose option B:binomcdf(. • Enter n, p, and x in the parentheses as: binomcdf (n, p, x). So, using this more efficient method, we would calculate the probability as shown below and in the screenshot in the margin.







Therefore, the probability that no more than two out of the fifteen batteries are defective is approximately 0.8159.

0.2

, ,

8159

P X x

X

n x

P

p

bi nomcdf ( 15, 0. 1,2)bi nomcdf ( )






Example 5.8: Calculating a Cumulative Binomial Probability Using a TI-83/84 Plus Calculator

What is the probability that a family with six children has more than two girls? Assume that the gender of one child is independent of the gender of any of the other children. SolutionFirst, let’s verify that this scenario meets the criteria of a binomial distribution. We are told that the gender of each child is independent, so we can consider the births of the children to be identical, independent trials.






Example 5.8: Calculating a Cumulative Binomial Probability Using a TI-83/84 Plus Calculator (cont.)

The family has six children, so the number of trials is n = 6. For each trial, there are two possible outcomes: the child is either a girl or a boy. Let’s define a success as having a girl. We can assume that both genders are equally likely; thus the probability of obtaining a success is p = 0.5. Let X = the number of girls out of the six children. We are considering the event of having more than two girls, X > 2. This is the complement to the event of having no more than two girls, X ≤ 2. Thus we can calculate the binomial probability by using the Complement Rule, as follows.







Using the binomial formula for this problem would be cumbersome, so let’s use a TI-83/84 Plus calculator. We would enter the probability expression as shown below and in the screenshot in the margin.

2 1 2P X P X

2 1 2

0.6563

P X P X

1Þbi nomcdf ( 6, 0. 5,2)







Therefore, the probability that a family with six children has more than two girls is approximately 0.6563.






Example 5.9: Calculating a Cumulative Binomial Probability Using a TI-83/84 Plus Calculator

Suppose that 20% of the programs sold at the home games of a professional sports team during the course of one season contain a special discount coupon. If all eight friends in your group bought programs at one game, what is the probability that at least half of your friends received the discount coupon?







SolutionSince there are a significant number of programs sold at the home games of a professional sports team throughout one season, and the number of trials we are considering is relatively small in comparison, we can model this situation as if the trials are independent. The reasoning is that the precise probabilities would not actually change enough to affect the value of the answer. We are considering eight programs; thus n = 8.







If we define a success to be receiving a discount coupon, then the probability of obtaining a success is p = 0.2. Let X = the number of discount coupons received in the eight programs bought by your friends. We are interested in the probability that at least half of the eight friends get a discount coupon, so at least four out of the eight, or As in the previous example, in order to use the cumulative binomial probability function on a TI-83/84 Plus calculator to solve this problem, we will need to use the Complement Rule.

4 .P X







This is still not exactly what we need because the TI-83/84 Plus calculator can only calculate cumulative probabilities of the form Fortunately, this situation is not too difficult to deal with due to one of the characteristics of the binomial distribution. The value for x must be a whole number; therefore,

4 1 4P X P X

.P X x

4 3 .P X P x







Using all of this information and a TI-83/84 Plus calculator, we calculate the probability as shown below and in the screenshot.

4 1

0.05

1

63

3

4

P

X P X

X

P

1Þbi nomcdf ( 8, 0. 2,3)







Therefore, the probability that at least half of the eight friends find discount coupons in their programs is approximately 0.0563, which indicates that it is not very likely.

Excel =BINOM.DIST.RANGE(n, p, xLow, xHigh)

Section 5.2

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