Binomial FormulasBinomial FormulasTarget Goal: I can calculate the Target Goal: I can calculate the mean and standard deviation of mean and standard deviation of

a binomial function.a binomial function.

6.3b6.3b

h.w: pg 404: 75, 77, 79 - 89h.w: pg 404: 75, 77, 79 - 89

1.1. Suppose I have a group of Suppose I have a group of 4 students4 students and I want to and I want to choose 1choose 1 of them as a of them as a volunteer. In how many ways can I volunteer. In how many ways can I choose 1 out of 4 students? choose 1 out of 4 students?

SSNNNNNN NNSSNNNN NNNNSSNN NNNNNNSS Call this Call this “4 choose 1.”“4 choose 1.” There are There are 4 4

waysways..

2.2. Suppose I have a group of Suppose I have a group of 4 students4 students and I want to and I want to choose 2choose 2 of them as of them as volunteers. In how many ways can I volunteers. In how many ways can I choose 2 out of 4 students? choose 2 out of 4 students?

SSSSNNNN SSNNSSNN SSNNNNSS NNSSSSNNNNSSNNSS NNNNSSSS

Call this Call this “4 choose 2.”“4 choose 2.” There are There are 6 6 ways.ways.

3.3. Suppose I have a group of Suppose I have a group of 5 5 studentsstudents and I want to and I want to choose 1choose 1 of of them as a volunteer. In how many them as a volunteer. In how many ways can I choose 1 out of 5 ways can I choose 1 out of 5 students?students?

SSNNNN NNNNN NSSNNN NNNNN NNSSNN NNNNN NNNSSN N NNNNNNNNSS

Call this Call this “5 choose 1.”“5 choose 1.” There are There are 5 5 waysways..

Binomial Coefficient Binomial Coefficient

There is a mathematical way to There is a mathematical way to countcount the total number of ways to the total number of ways to arrange k out of narrange k out of n objects. This is objects. This is called called “n choose k”“n choose k” or the or the binomial binomial coefficient.coefficient.

Binomial Coefficient:Binomial Coefficient: the number of the number of ways to ways to arrange k successes in n arrange k successes in n observationsobservations. .

It is written and is called It is written and is called “n “n choose k.”choose k.”

The value of “n choose k” is given by The value of “n choose k” is given by the formula:the formula:

n

k

!

! !

n n

k k n k

Example: “5 choose 2”Example: “5 choose 2”

5!

2! 3 !

5 4 3 2 1

2 1 3 2 1

5 5!

2 2! 5 2 !

10

So, there are So, there are 10 ways 10 ways to to arrange arrange

2 out of 5 objects.2 out of 5 objects. Think of this as flipping a coin 5 Think of this as flipping a coin 5

times and getting 2 heads. times and getting 2 heads.

Ex. Inheriting Blood Type Ex. Inheriting Blood Type

Recall:Recall: p = 0.25p = 0.25, probability child has type O , probability child has type O

bloodblood n = 5n = 5, family has 5 children, family has 5 children

Find the probability Find the probability exactly 2 exactly 2 childrenchildren have type O. have type O.

Which 2 children?Which 2 children?

Step 1: Step 1:

Find the probability Find the probability child #1child #1 and and #3 #3 havehave type O blood. type O blood.

SS F F SS F F F F

Prob:Prob: (.25)(.25)(.75)(.75)(.25)(.25)(.75)(.75)(.75)(.75)

= = (.25)(.25)22(.75)(.75)33

Step 2:Step 2:

All possible arrangements of 2 All possible arrangements of 2 successes and 3 failures will give successes and 3 failures will give (.25)(.25)22(.75)(.75)33..

How many arrangements? How many arrangements?

5 choose 2 = 5 choose 2 = 1010 So, P(X=2So, P(X=2) = # arrangements) = # arrangements xx prob. prob.

= 10= 10 (.25)(.25)22(.75)(.75)33

= .2637= .2637

Recall: n! = n x (n-1) x (n-2) x … x 3 x 2 x 1Recall: n! = n x (n-1) x (n-2) x … x 3 x 2 x 1 0! = 10! = 1

Binomial ProbabilityBinomial ProbabilityIf X has the If X has the binomial distributionbinomial distribution with with n observations and probability p and n observations and probability p and k is one of the possible values then,k is one of the possible values then,

P(X = k) = P(X = k) = ppkk(1-p)(1-p)n-kn-k

(# arrangements (# arrangements xx prob) prob)

1010 (.25)(.25)22(.75)(.75)33

n

k

Ex. Defective SwitchesEx. Defective Switches (no calculators) (no calculators)

n = 10n = 10, , p = 0.1p = 0.1, , XX is the is the # of switches that # of switches that failfail

Find the probability Find the probability no more than 1no more than 1 switch switch fails.fails.

P(X ≤ 1)P(X ≤ 1) = P(X = 0) + P(X = 1) = P(X = 0) + P(X = 1)

= = (.10)(.10)00(.9)(.9)10 10 + + (.10)(.10)11(.9)(.9)9 9

==

= 0.3487 + 0.3874= 0.3487 + 0.3874

= = 0.73610.7361

10

1

10

0

11 0.3487 .1 0.3874

0! 10!

0!10! 1!9!

Binomial Mean and Standard Binomial Mean and Standard DeviationDeviation

If X is a If X is a binomial random variablebinomial random variable with with parameters n and p, then the parameters n and p, then the meanmean and and standard deviationstandard deviation of X are: of X are:

Or,Or,

(Only for binomial, not for discrete random (Only for binomial, not for discrete random variables.)variables.)

1

X

X

np

np p

x npq

Ex: Bad SwitchesEx: Bad Switches

The count of bad switches with The count of bad switches with n = n = 1010 and and p = 0.1p = 0.1 (bad switches). (bad switches).

This is the This is the sampling distributionsampling distribution an an engineerengineer would see if would see if she drew all she drew all possible SRS’s of 10 switchespossible SRS’s of 10 switches from a from a shipment and recorded the value of shipment and recorded the value of X for each sample.X for each sample.

Find the Find the mean and standard deviationmean and standard deviation of of the distribution.the distribution.

σσ = 0.9487 = 0.9487

(10)(0.1)(0.9) 0.

(10)(0.1) 1

9

x

x

1X

X np

np p

The Normal Approximation to The Normal Approximation to Binomial DistributionsBinomial Distributions

As the As the number of trials n gets largernumber of trials n gets larger, , the binomial distribution X the binomial distribution X gets close gets close to a normal distribution.to a normal distribution.

( , (1 ))N np np p

The probability histogram for the The probability histogram for the binomial distribution:binomial distribution:

As a As a ”rule of thumb””rule of thumb”, we use the , we use the normal approx.when n and p satisfy:normal approx.when n and p satisfy:

np ≥ 10np ≥ 10

n(1-p) ≥ 10n(1-p) ≥ 10 s/a nq ≥ s/a nq ≥ 1010

Ex: Attitudes Toward Shopping Ex: Attitudes Toward Shopping

A nationwide random sample A nationwide random sample surveyed surveyed 2500 adults2500 adults about about shopping. Suppose that in fact shopping. Suppose that in fact 60% 60% of all adult U.Sof all adult U.S. residents would say . residents would say they agree that they agree that “they like buying “they like buying new cloths, but shopping is often new cloths, but shopping is often frustrating and time consuming.”frustrating and time consuming.”

What is the probability that 1520 or What is the probability that 1520 or more of the sample agree?more of the sample agree?

Check:Check: There are There are 195 million adults195 million adults, so , so

sample is independent.sample is independent.

npnp = = 2500(0.6)2500(0.6) = 1500 = 1500 ≥ 10≥ 10

n(1-p)n(1-p) = = 2500(0.4)2500(0.4) = 1000 = 1000 ≥ 10≥ 10

So, we can assume N(μ,σ)So, we can assume N(μ,σ)

μ = npμ = np = = 15001500

σ = σ =

= =

= 24.49= 24.49

npq

(2500)(0.6)(0.4)

A normal distribution approximates A normal distribution approximates our binomial distribution well. So do our binomial distribution well. So do a a normal calculation.normal calculation.

P(X≥1520) = P(X≥1520) =

= = P(Z≥0.82)P(Z≥0.82)

= 1 – 0.7939 or = 1 – 0.7939 or normcdf(.82,EE99)normcdf(.82,EE99)

= = 0.20610.2061

The probability that The probability that 1520 or more1520 or more of of the sample the sample agreeagree is is approximately approximately 20.61%.20.61%.

1520 15001520

24.49P X P Z

Exercise 8.15: Attitudes on Exercise 8.15: Attitudes on ShoppingShopping

b. Use your calculator and the b. Use your calculator and the cumulative binomial function to cumulative binomial function to verifyverify the exact answer for the the exact answer for the probability probability at least 1520 people in at least 1520 people in the sample find shopping frustrating the sample find shopping frustrating is 0.2131. is 0.2131.

What is the probability to 6 decimal What is the probability to 6 decimal places?places?

P(XP(X≥≥1520) = 1520) = 1 – P(X≤1519)1 – P(X≤1519)

= 1 - = 1 - binomcdf(2500,0.6,1519)binomcdf(2500,0.6,1519)

= 1 - .7868609113= 1 - .7868609113

= = .213139.213139

What is the probability that What is the probability that at most 1468at most 1468 people in the sample would agree with the people in the sample would agree with the statement that shopping is frustrating?statement that shopping is frustrating?

Standardize:Standardize:

= P(Z≤ -1.29)= P(Z≤ -1.29)

normcdf(-EE99,-1.29)normcdf(-EE99,-1.29)

= = .0985.0985

1468 1500

24.1468

49P P ZX