joyful mood is a meritorious deed that cheers up people around you

1

Joyful mood is a meritorious deed that cheers up people around you

like the showering of cool spring breeze.

BPS - 5th Ed. Chapter 13 2

Chapter 13

Binomial Distributions


Fixed number n of observations The n observations are independent Each observation falls into one of just

two categories– may be labeled “success” and “failure”

The probability of success, p, is the same for each observation

Binomial Setting


In a shipment of 100 televisions, how many are defective?– counting the number of “successes”

(defective televisions) out of 100 A new procedure for treating breast

cancer is tried on 25 patients; how many patients are cured?– counting the number of “successes”

(cured patients) out of 25

Binomial SettingExamples


Let X = the count of successes in a binomial setting. The distribution of X is the binomial distribution with parameters n and p.– n is the number of observations– p is the probability of a success on any

one observation– X takes on whole values between 0 and n

Binomial Distribution


– not all counts have binomial distributions trials (observations) must be independent the probability of success, p, must be the same

for each observation– if the population size is MUCH larger than the

sample size n, then even when the observations are not independent and p changes from one observation to the next, the change in p may be so small that the count of successes (X) has approximately the binomial distribution

Binomial Distribution


Case StudyInspecting Switches

An engineer selects a random sample of 10 switches from a shipment of 10,000 switches. Unknown to the engineer, 10% of the switches in the full shipment are bad. The engineer counts the number X of bad switches in the sample.



X (the number of bad switches) is not quite binomial– Removing one switch changes the proportion of bad

switches remaining in the shipment (selections are not independent)

However, removing one switch from a shipment of 10,000 changes the makeup of the remaining 9,999 very little– the distribution of X is very close to the binomial

distribution with n=10 and p=0.1


Find the probability that a binomial random variable takes any particular value– P(x successes out of n observations) = ?– need to add the probabilities for the

different ways of getting exactly x successes in n observations

Binomial Probabilities


Each offspring hatched from a particular type of reptile has probability 0.2 of surviving for at least one week. If 6 offspring of these reptiles are hatched, find the probability that exactly 2 of the 6 will survive for at least one week. Label an offspring that survives with S for “success” and one that dies with F for “failure”.P(S) = 0.2 and P(F) = 0.8.

Binomial ProbabilitiesExample


(1) First, find probability that the two survivors are the first two offspring:

Using the Multiplication Rule:

P(SSFFFF) = (0.2)(0.2)(0.8)(0.8)(0.8)(0.8) = (0.2)2(0.8)4

= 0.0164



(2) Second, find the number of possible arrangements for getting two successes and four failures:SSFFFF SFSFFF SFFSFF SFFFSF SFFFFSFSSFFF FSFSFF FSFFSF FSFFFS FFSSFFFFSFSF FFSFFS FFFSSF FFFSFS FFFFSS


There are 15 of these, and each has the same probability of occurring: (0.2)2(0.8)4.

Thus, the probability of observing exactly 2 successes out of 6 is: P(X=2) = 15(0.2)2(0.8)4 = 0.246 .


The number of ways of arranging k successes among n observations is given by the binomial coefficient

Binomial Coefficient

)!(!!

knkn

kn

where n! is “n factorial” (see next slide).– the binomial coefficient is read “n choose k”.


For any positive whole number n, its factorial n! is

Factorial Notation

n! = n (n1) (n2) 3 2 1– Also, 0! = 1 by definition.

Example: 6! = 6·5·4·3·2·1 = 720,and from the previous example:

152

301256

1)2341)(2123456

2!4!6!

2)!(62!6!

26

(


If X has the binomial distribution with n observations and probability p of success on each observation, the possible values of X are 0, 1, 2, …, n. If k is any one of these values, then

Binomial Probabilities

knk ppkn

kX

)(1)P(



The number X of bad switches has approximately the binomial distribution with n=10 and p=0.1. Find the probability of getting 1 or 2 bad switches in a sample of 10.

0.5811

0.19370.3874(0.4305)(45)(0.01)0.3874)(10)(0.1)(

(0.9)(0.1)2!8!10!(0.9)(0.1)

1!9!10!

0.1)(1-(0.1)2100.1)(1-(0.1)1

102)P(1)P(2) or 1P(

8291

2-1021-101

XXX


If X has the binomial distribution with n observations and probability p of success on each observation, then the mean and standard deviation of X are

Mean and Standard Deviation

)(1σ

μ

pnp

np



The number X of bad switches has approximately the binomial distribution with n=10 and p=0.1. Find the mean and standard deviation of this distribution.

µ = np = (10)(0.1) = 1 the probability of each being bad is one tenth; so we

expect (on average) to get 1 bad one out of the 10 sampled

0.94870.90.1)1(10)(0.1)()(1σ pnp



Probability Histogramn=10, p=0.1


The formula for binomial probabilities becomes cumbersome as the number of trials n increases

As the number of trials n increases, the binomial distribution gets close to a Normal distribution– when n is large, Normal probability

calculations can be used to approximate binomial probabilities

Normal Approximationto the Binomial


The Normal distribution that is used to approximate the binomial distribution uses the same mean and standard deviation:

Normal Approximationto the Binomial

)(1σ and μ pnpnp When n is large, a binomial random variable

X (with n trials and success probability p) is approximately Normal:

)(1 , N approx. is pnpnpX


As a rule of thumb, we will use the Normal approximation to the Binomial when n is large enough to satisfy the following:

np ≥ 10 and n(1p) ≥ 10– Note that these conditions also depend

on the value of p (and not just on n)

Normal Approximationto the Binomial (Sample Size)


Nationwide random sample of 2500 adults were asked if they agreed or disagreed with the statement “I like buying clothes, but shopping is often frustrating and time-consuming.” Suppose that in fact 60% of the population of all adult U.S. residents would say “Agree” if asked this question. What is the probability that 1520 or more of the sample agree?

Case StudyShopping Attitudes

Hall, Trish. “Shop? Many say ‘Only if I must’,” New York Times, November 28, 1990.


Shopping Attitudes

The responses of the 2500 randomly chosen adults (from over 210 million adults) can be taken to be independent.

The number X in the sample who agree has a binomial distribution with n=2500 and p=0.60.

To find the probability that at least 1520 people in the sample agree, we would need to add the binomial probabilities of all outcomes from X=1520 to X=2500…this is not practical.

Case Study


Histogram of 1000 simulated values of the binomial variable X, and the density curve of the Normal distribution with the same mean and standard deviation:

Case StudyShopping Attitudes

µ = np = 2500(0.6) = 1500

24.49600

)(0.4)(2500)(0.6

)(1σ

pnp

Find probability of getting at least 1520:


Shopping Attitudes

Assuming X has the N(1500, 24.49) distribution [np and n(1p) are both ≥ 10], we have

Case Study

0.20610.79391

0.82)P(Z24.49

15001520σ

μP1520)P(

Table) Normal Standard (from

XX


Shopping Attitudes

The probability of observing 1520 or more adults in the sample who agree with the statement has been calculated as 20.61% using the Normal approximation to the Binomial.

Using a computer program to calculate the actual Binomial probabilities for all values from 1520 to 2500, the true probability of observing 1520 or more who agree is 21.31%

This is a very good approximation!

Case Study

joyful mood is a meritorious deed that cheers up people around you

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