THE BINOMIAL DISTRIBUTION& ITS APPLICATIONS

Presented By: Umar Farooq, Umair Javed, Shamim Aslam, Hamza Akash & ,Haseeb Hayat

BINOMIAL DISTRIBUTIONThe main objective of this is to cover the basics of binomial distribution, study some examples and look at its Advantages and Disadvantages.

BASICSBefore we begin new content, we should review a few terms from previous lessons that

we will see again inthis lesson: Discrete: Data that can only take on set number of values Continuous: Quantitative data that can take on any value between the minimum

and maximum, and any value between two other values Probability: The likelihood of an event occuring ; P(A)=number of events

considered outcome A number of total eventsP(A) = number of events considered out come A number of total events

P(A∩B)P(A∩B): Intersection of A and B; "probability of A and B" P(A∪B)P(A∪B): Union of A and B; "probability of A or B" (this also includes the

probability of A and B) Mean: The numerical average calculated as the sum of all of the data values

divided by the number of values; represented as X¯¯¯¯X¯. Standard deviation: Roughly the average difference between individual data and

the mean; for a sample, represented as s, s=∑(x−x¯¯¯)2n−1−−−−−−√s=∑(x−x¯)2n−1

Sample: A subset of the population from which data is actually collected Population: The entire set of possible observations in which we are interested Statistic: A measure concerning a sample (e.g., sample mean) Parameter: A measure concerning a population (e.g., population mean)

PROBABILITY DISTRIBUTION PREREQUISITESTo understand probability distributions, it isimportant to understand variables. Randomvariables, and some notation. A variable is a symbol (A, B, x, y, etc.) that can take on any of

a specified set of values. When the value of a variable is the outcome of a 

statistical experiment, that variable is a random variable.Generally, statisticians use a capital letter to represent a randomvariable anda lower-case letter, to represent one of its values. For example, X represents the random variable X. P(X) represents the probability of X. P(X = x) refers to the probability that the random variable X is

equal to a particular value, denoted by x. As an example, P(X = 1) refers to the probability that the random variable X is equal to 1.

RANDOM VARIABLEThe mathematical rule (or function) that

assigns a given numerical value to each possible outcome of an experiment in the sample space of interest.

Discrete random variables Continuous random variables

WHAT ARE PROBABILITY DISTRIBUTIONS?A probability distribution is a table or an

equation that links each outcome of a statistical experiment with its probability of occurrence.

A discrete probability distribution specifies:the possible values of the random variable,

andthe probability that each outcome will occur

PROBABILTY DISTRIBUTIONS An example will make clear the relationship between random

variables and probability distributions. Suppose you flip a coin two times. This simple statistical experiment can have four possible outcomes: HH, HT, TH, and TT. Now, let the variable X represent the number of Heads that result from this experiment. The variable X can take on the values 0, 1, or 2. In this example, X is a random variable; because its value is determined by the outcome of a statistical experiment.

A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. Consider the coin flip experiment described above. The table below, which associates each outcome with its probability, is an example of a probability distribution.Number of Heads Probability

0 0.251 0.502 0.25

Distribution Table

DISTRIBUTIONSHere are some of the common distributions and some of the reasons that they are

useful: Normal: This is useful for looking at means and other linear combinations (e.g.

regression coefficients) because of the CLT. Related to that is if something is known to arise due to additive effects of many different small causes then the normal may be a reasonable distribution: for example, many biological measures are the result of multiple genes and multiple environmental factors and therefor are often approximately normal.

Gamma: Right skewed and useful for things with a natural minimum at 0. Commonly used for elapsed times and some financial variables.

Exponential: special case of the Gamma. It is memoryless and scales easily. Chi-squared (χ2χ2): special case of the Gamma. Arises as sum of squared normal

variables (so used for variances). Beta: Defined between 0 and 1 (but could be transformed to be between other

values), useful for proportions or other quantities that must be between 0 and 1. Binomial: How many "successes" out of a given number of independent trials with

same probability of "success". Poisson: Common for counts. Nice properties that if the number of events in a period

of time or area follows a Poisson, then the number in twice the time or area still follows the Poisson (with twice the mean): this works for adding Poissons or scaling with values other than 2.

You could use a tree diagram

GRAPHS OF SOME DISTRIBUTIONS

PROBABILITY DISTRIBUTIONS We use probability

distributions because they work –they fit lots of data in real world

TRY THISHave you got a coin?

Toss it six times in a roll, each time counting

the number of times the result, ‘heads’ isobserved.

BINOMIAL EXPERIMENT There are many probability

experiments for which the results of each trial can be reduced to two outcomes: success and failure.

For instance, when a basketball player attempts a free throw, he or she either makes the basket or does not. Probability experiments such as these are called binomial experiments.

OBSERVED PROBABILITYBefore a coin is tossed six times in a roll,what is the probability that in total there

willbe two ‘heads’ out of six?

BERNOULLI TRIALTossing a coin is Bernoulli trial. A Bernoulli trial is arandom experiment that has only two, mutuallyexclusive outcomes.

Thus, when a coin is tossed, the two possible outcomes are revealing a ‘heads’ or revealing a ‘tails’. We want to see how many ‘heads’ are revealed. So revealing a ‘heads’ is a ‘success’. Revealing a ‘tails’ is a ‘failure’.

PROPERTIES OF A BERNOULLI TRIAL

Since you’re tossing the same coin in all sixBernoulli trials, the probability of a ‘heads’ or asuccess, p, is the same for each repeat of theBernoulli trial

But the coin has no memory: Bernoulli trials are

independent

MULTIPLE ANSWER QUESTION: WHICH OF THEFOLLOWING ARE REPEATS OF BERNOULLI TRIALS

1.The game of joker’s challenge is played withthe full deck of cards: The player is shown a card,and s/he calls whether the next card will be loweror higher, if the call is correct the game isrepeated, and so on2.A box contains 20 packs of chocolate, 8 ofwhich are milk. Selecting a chocolate, checking ifits milk, eating eat it, really enjoying it and thenrepeating the exercise.

BINOMIAL DISTRIBUTION"Bi" means "two"  (like a bicycle has two

wheels) ... 

... so this is about things with two results.  

FLIPPING A COINTossing a Coin:Did we get Heads (H)

or

Tails (T)We say the probability of the coin landing H is ½ And the probability of the coin landing T is ½

THROWING A DIEThrowing a Die: Did we get a four ... ? ... or not? We say the probability of a four is 1/6

(one of the six faces is a four).And the probability of not four is 5/6 (five of the six faces are not a four)

DEFINATION A binomial experiment is a probability experiment that

satisfies the following conditions:

1. The experiment is repeated for a fixed number of trials, where each trial is independent of the other trials.

2. There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success (S) or as a failure (F).

3. The probability of a success, P(S), is the same for each trial.

4. The random variable, x, counts the number of successful trials

NOTATION FOR BINOMIAL EXPERIMENTS

Symbol Description

n The number of times a trial is repeated.

p = P(S) The probability of success in a single trial.

q = P(F) The probability of failure in a single trial (q = 1 – p)

x The random variable represents a count of the number of successes in n trials: x = 0, 1, 2, 3, . . . n.

HOW TO DETERMINE? Decide whether the experiment is a binomial

experiment. If it is, specify the values of n, p and q and list the possible values of the random variable, x. If it is not, explain why.

A certain surgical procedure has an 85% chance of success. A doctor performs the procedure on eight patients. The random variable represents the number of successful surgeries.

BINOMIAL EXPERIMENTSSolution: the experiment is a binomial experiment

because it satisfies the four conditions of a binomial experiment. In the experiment, each surgery represents one trial. There are eight surgeries, and each surgery is independent of the others. Also, there are only two possible outcomes for each surgery—either the surgery is a success or it is a failure. Finally, the probability of success for each surgery is 0.85.

n = 8p = 0.85q = 1 – 0.85 = 0.15x = 0, 1, 2, 3, 4, 5, 6, 7, 8

BINOMIAL EXPERIMENTS2. A jar contains five red marbles, nine

blue marbles and six green marbles. You randomly select three marbles from the jar, without replacement. The random variable represents the number of red marbles.

BINOMIAL EXPERIMENTSSolution: The experiment is not a binomial

experiment because it does not satisfy all four conditions of a binomial experiment. In the experiment, each marble selection represents one trial and selecting a red marble is a success. When selecting the first marble, the probability of success is 5/20. However because the marble is not replaced, the probability is no longer 5/20. So the trials are not independent, and the probability of a success is not the same for each trial.

BINOMIAL EXAMPLE

Take the example of 5 coin tosses. What’s the probability that you flip exactly 3 heads in 5 coin tosses?

LET’S TOSS A COIN

HEAD HEAD HEADTAIL HEAD HEADHEAD HEAD TAILHEAD TAIL HEADHEAD TAIL TAILTAIL HEAD TAILTAIL TAIL HEADTAIL TAIL TAIL

Toss a fair coin threetimes ...what is the chance ofgetting two Heads?Tossing a coin three times(H is for heads, T for Tails) Can get any of these8 outcomes:

Which outcomes do we want?"Two Heads" could be in any order: "HHT", "THH" and "HTH"

allhave two Heads (and one Tail). So 3 of the outcomes produce "Two Heads".What is the probability of each outcome? Each outcome is equally likely, and there are 8 of them. So

each has a probability of 1/8So the probability of event "Two Heads" is:Number of outcomes we want * Probability of each outcome  3  ×  1/8  =  3/8

LETS CALCULATE THEM ALLThe calculations are (P means "Probability of"): P(Three Heads) = P(HHH) = 1/8 P(Two Heads) = P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8 P(One Head) = P(HTT) + P(THT) + P(TTH) = 1/8 + 1/8 + 1/8 = 3/8 P(Zero Heads) = P(TTT) = 1/8We can write this in terms of a Random Variable, X, = "The number ofHeads from 3 tosses of a coin":

P(X = 3) = 1/8 P(X = 2) = 3/8 P(X = 1) = 3/8 P(X = 0) = 1/8

THE BINOMIAL DISTRIBUTIONBERNOULLI RANDOM VARIABLES Imagine a simple trial with only two possible

outcomes Success (S) Failure (F)

Examples Toss of a coin (heads or tails) Sex of a newborn (male or female) Survival of an organism in a region (live or

die)

Jacob Bernoulli (1654-1705)

THE BINOMIAL DISTRIBUTIONOVERVIEW

Suppose that the probability of success is p

What is the probability of failure? q = 1 – p

Examples Toss of a coin (S = head): p = 0.5 q = 0.5 Roll of a die (S = 1): p = 0.1667 q = 0.8333 Fertility of a chicken egg (S = fertile): p = 0.8

q = 0.2

THE BINOMIAL DISTRIBUTIONOVERVIEW

Imagine that a trial is repeated n times

Examples A coin is tossed 5 times A die is rolled 25 times 50 chicken eggs are examined

Assume p remains constant from trial to trial and that the trials are statistically independent of each other

THE BINOMIAL DISTRIBUTIONOVERVIEW

What is the probability of obtaining x successes in n trials?

Example What is the probability of obtaining 2 heads

from a coin that was tossed 5 times?

P(HHTTT) = (1/2)5 = 1/32

THE BINOMIAL DISTRIBUTIONOVERVIEW

But there are more possibilities:

HHTTT HTHTT HTTHT HTTTHTHHTT THTHT THTTH

TTHHT TTHTHTTTHH

P(2 heads) = 10 × 1/32 = 10/32

THE BINOMIAL DISTRIBUTIONOVERVIEW

In general, if trials result in a series of success and failures,

FFSFFFFSFSFSSFFFFFSF…

Then the probability of x successes in that order is

P(x) = q q p q = px qn – x

THE BINOMIAL DISTRIBUTIONOVERVIEW

However, if order is not important, then

where is the number of ways to obtain x successes

in n trials, and i! = i (i – 1) (i – 2) … 2 1

n!x!(n – x)!

px qn – xP(x) =

n!x!(n – x)!

EXAMPLE As voters exit the polls, you ask a representative random

sample of 6 voters if they voted for proposition 100. If the true percentage of voters who vote for the proposition is 55.1%, what is the probability that, in your sample, exactly 2 voted for the proposition and 4 did not?

 

SOLUTION:

Outcome Probability YYNNNN = (.551)2 x (.449)4

NYYNNN (.449)1 x (.551)2 x (.449)3 = (.551)2 x (.449)4

NNYYNN (.449)2 x (.551)2 x (.449)2 = (.551)2 x (.449)4

NNNYYN (.449)3 x (.551)2 x (.449)1 = (.551)2 x (.449)4

NNNNYY (.449)4 x (.551)2 = (.551)2 x (.449)4

.

.

ways to arrange 2 Obama votes among 6 voters

6

2

15 arrangements x (.551)2 x (.449)4  

6

2P(2 yes votes exactly) = x (.551)2 x (.449)4 = 18.5%

BINOMIAL PROBABILITIESThere are several ways to find the probability of x

successes in n trials of a binomial experiment. One way is to use the binomial probability formula.

Binomial Probability FormulaIn a binomial experiment, the probability of exactly x

successes in n trials is:

xnxxnxxn qp

xxnnqpCxP

!)!(!)(

BINOMIAL EXAMPLE

Take the example of 5 coin tosses. What’s the probability that you flip exactly 3 heads in 5 coin tosses?

BINOMIAL DISTRIBUTION

Solution:One way to get exactly 3 heads: HHHTT

What’s the probability of this exact arrangement?

P(heads)xP(heads) xP(heads)xP(tails)xP(tails) =(1/2)3 x (1/2)2

Another way to get exactly 3 heads: THHHT

Probability of this exact outcome = (1/2)1 x (1/2)3 x (1/2)1 = (1/2)3 x (1/2)2

BINOMIAL DISTRIBUTION

In fact, (1/2)3 x (1/2)2 is the probability of each unique outcome that has exactly 3 heads and 2 tails.

So, the overall probability of 3 heads and 2 tails is:(1/2)3 x (1/2)2 + (1/2)3 x (1/2)2 + (1/2)3 x (1/2)2 + ….. for as many unique arrangements as there are—but how many are there??

 

Outcome Probability THHHT (1/2)3 x (1/2)2

HHHTT (1/2)3 x (1/2)2

TTHHH (1/2)3 x (1/2)2

HTTHH (1/2)3 x (1/2)2

HHTTH (1/2)3 x (1/2)2

THTHH (1/2)3 x (1/2)2

HTHTH (1/2)3 x (1/2)2

HHTHT (1/2)3 x (1/2)2

THHTH (1/2)3 x (1/2)2

HTHHT (1/2)3 x (1/2)2

10 arrangements x (1/2)3 x (1/2)2

 

The probability of each unique outcome (note: they are all equal)

ways to arrange 3 heads in 5 trials

5

3

5C3 = 5!/3!2! = 10

 P(3 heads and 2 tails) = x P(heads)3 x P(tails)2 =

10 x (½)5=31.25%

5

3

x

p(x)

0 3 4 51 2

BINOMIAL DISTRIBUTION FUNCTION:X= THE NUMBER OF HEADS TOSSED IN 5 COIN TOSSES

number of heads

p(x)

number of heads

BINOMIAL DISTRIBUTION, GENERALLY

XnXn

Xpp

)1(

1-p = probability of failure

p = probability of success

X = # successes out of n trials

n = number of trials

Note the general pattern emerging if you have only two possible outcomes (call them 1/0 or yes/no or success/failure) in n independent trials, then the probability of exactly X “successes”=

FINDING BINOMIAL PROBABILITIESA six sided die is rolled 3 times. Find the probability

of rolling exactly one 6.Roll 1 Roll

2Roll 3

Frequency # of 6’s Probability

(1)(1)(1) = 1 3 1/216

(1)(1)(5) = 5 2 5/216

(1)(5)(1) = 5 2 5/216

(1)(5)(5) = 25 1 25/216

(5)(1)(1) = 5 2 5/216

(5)(1)(5) = 25 1 25/216

(5)(5)(1) = 25 1 25/216

(5)(5)(5) = 125 0 125/216

You could use a tree diagram

FINDING BINOMIAL PROBABILITIESThere are three outcomes that have exactly one six,

and each has a probability of 25/216. So, the probability of rolling exactly one six is 3(25/216) ≈ 0.347. Another way to answer the question is to use the binomial probability formula. In this binomial experiment, rolling a 6 is a success while rolling any other number is a failure. The values for n, p, q, and x are n = 3, p = 1/6, q = 5/6 and x = 1. The probability of rolling exactly one 6 is:

xnxxnxxn qp

xxnnqpCxP

!)!(!)(

Or you could use the binomial probability formula

FINDING BINOMIAL PROBABILITIES

347.07225

)21625(3

)3625)(

61(3

)65)(

61(3

)65()

61(

!1)!13(!3)1(

2

131

P

By listing the possible values of x with the corresponding probability of each, you can construct a binomial probability distribution.

EXAMPLE:A fair die is thrown four times. Calculate the probabilities of getting: 0 Twos 1 Two 2 Twos 3 Twos 4 TwosIn this case n=4, p = P(Two) = 1/6X is the Random Variable ‘Number of Twos from four throws’.Substitute x = 0 to 4 into the formula: P(k out of n) =  {n! /k!(n-k )! }pk(1-p)(n-k) Like this (to 4 decimal places): P(X = 0) = (4!/0!4!) × (1/6)0(5/6)4 = 1 × 1 × (5/6)4 = 0.4823 P(X = 1) = (4!/1!3!) × (1/6)1(5/6)3 = 4 × (1/6) × (5/6)3 = 0.3858 P(X = 2) = (4!/2!2!) × (1/6)2(5/6)2 = 6 × (1/6)2 × (5/6)2 = 0.1157 P(X = 3) = (4!/3!1!) × (1/6)3(5/6)1 = 4 × (1/6)3 × (5/6) = 0.0154 P(X = 4) = (4!/4!0!) × (1/6)4(5/6)0 = 1 × (1/6)4 × 1 = 0.0008Summary: "for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12%chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% chance of all throws being atwo (but it still could happen!)“This time the Bar Graph is not symmetrical:

EXAMPLE 2:Your company makes sports bikes. 90% pass final inspection (and 10% fail

and need to be fixed).What is the expected Mean and Variance of the 4 next inspections? First, let's calculate all probabilities. n = 4, p = P(Pass) = 0.9 X is the Random Variable "Number of passes from four inspections". Substitute x = 0 to 4 into the formula: P(k out of n) =  {n!/ k!(n-k)!} pk(1-p)(n-kLike this: P(X = 0) = (4!/0!4!) × 0.900.14 = 1 × 1 × 0.0001 = 0.0001 P(X = 1) = (4!/1!3!) × 0.910.13 = 4 × 0.9 × 0.001 = 0.0036 P(X = 2) = (4!/2!2!) × 0.920.12 = 6 × 0.81 × 0.01 = 0.0486 P(X = 3) = (4!/3!1!) × 0.930.11 = 4 × 0.729 × 0.1 = 0.2916 P(X = 4) = (4!/4!0!) × 0.940.10 = 1 × 0.6561 × 1 = 0.6561 Summary: "for the 4 next bikes, there is a tiny 0.01% chance of no

passes, 0.36% chance of 1 pass, 5% chance of 2 passes, 29% chance of 3 passes, and a whopping 66% chance they all pass the inspection."

FINDING BINOMIAL PROBABILITIES

A survey indicates that 41% of American women consider reading as their favorite leisure time activity. You randomly select four women and ask them if reading is their favorite leisure-time activity. Find the probability that (1) exactly two of them respond yes, (2) at least two of them respond yes, and (3) fewer than two of them respond yes.

FINDING BINOMIAL PROBABILITIES #1--Using n = 4, p = 0.41, q = 0.59 and x =2, the

probability that exactly two women will respond yes is:

35109366.)3481)(.1681(.6

)3481)(.1681(.424

)59.0()41.0(!2)!24(

!4)59.0()41.0()2(

242

24224

CP

P(2) exactly two women

FINDING BINOMIAL PROBABILITIES #2--To find the probability that at least two women will respond

yes, you can find the sum of P(2), P(3), and P(4). Using n = 4, p = 0.41, q = 0.59 and x =2, the probability that at least two women will respond yes is:

028258.0)59.0()41.0()4(

162653.0)59.0()41.0()3(

351093.)59.0()41.0()2(

44444

34334

24224

CP

CP

CP

542.0028258162653.351093.

)4()3()2()2(

PPPxP

Use Calculator

FINDING BINOMIAL PROBABILITIES #3--To find the probability that fewer than two women will respond

yes, you can find the sum of P(0) and P(1). Using n = 4, p = 0.41, q = 0.59 and x =2, the probability that at least two women will respond yes is:

336822.0)59.0()41.0()1(

121174.0)59.0()41.0()0(141

14

04004

CP

CP

458.0336822.121174..

)1()0()2(

PPxP

For fewer than two women

NOTE:

Finding binomial probabilities with the binomial formula can be a tedious and mistake prone process. To make this process easier, you can use a binomial probability Table that lists the binomial probability for selected values of n and p.

FINDING A BINOMIAL PROBABILITY USING A TABLE Fifty percent of working adults spend less than 20 minutes

commuting to their jobs. If you randomly select six working adults, what is the probability that exactly three of them spend less than 20 minutes commuting to work? Use a table to find the probability.

Solution: A portion of Table 2 is shown here. Using the distribution for n = 6 and p = 0.5, you can find the probability that x = 3, as shown by the highlighted areas in the table.

In a survey, American workers and retirees are askConstructing a Binomial Distribution

ed to name their expected sources of retirement income. The results are shown in the graph. Seven workers who participated in the survey are asked whether they expect to rely on social security for retirement income. Create a binomial probability distribution for the number of workers who respond yes.

SOLUTION you can see that 36% of working Americans expect to rely on

social security for retirement income. So, p = 0.36 and q = 0.64. Because n = 7, the possible values for x are 0, 1, 2, 3, 4, 5, 6 and 7.

044.0)64.0()36.0()0( 7007 CP

173.0)64.0()36.0()1( 6117 CP

292.0)64.0()36.0()2( 5227 CP

274.0)64.0()36.0()3( 4337 CP

154.0)64.0()36.0()4( 3447 CP

052.0)64.0()36.0()5( 2557 CP

010.0)64.0()36.0()6( 1667 CP

001.0)64.0()36.0()7( 0777 CP

x P(x)

0 0.044

1 0.173

2 0.292

3 0.274

4 0.154

5 0.052

6 0.010

7 0.001

P(x) = 1

Notice all the probabilities are between 0 and 1 and that the sum of the probabilities is 1.

CONSTRUCTING AND GRAPHING A BINOMIAL DISTRIBUTION

65% of American households subscribe to cable TV. You randomly select six households and ask each if they subscribe to cable TV. Construct a probability distribution for the random variable, x. Then graph the distribution.

Calculator or look it up on pg. A10075.0)35.0()65.0()6(

244.0)35.0()65.0()5(

328.0)35.0()65.0()4(

235.0)35.0()65.0()3(

095.0)35.0()65.0()2(

020.0)35.0()65.0()1(

002.0)35.0()65.0()0(

66666

56556

46446

36336

26226

16116

06006

CP

CP

CP

CP

CP

CP

CP

CONSTRUCTING AND GRAPHING A BINOMIAL DISTRIBUTION

65% of American households subscribe to cable TV. You randomly select six households and ask each if they subscribe to cable TV. Construct a probability distribution for the random variable, x. Then graph the distribution.

Because each probability is a relative frequency, you can graph the probability using a relative frequency histogram as shown on the next slide.

x 0 1 2 3 4 5 6

P(x) 0.002 0.020 0.095 0.235 0.328 0.244 0.075

CONSTRUCTING AND GRAPHING A BINOMIAL DISTRIBUTION Then graph the distribution.

x 0 1 2 3 4 5 6P(x) 0.002 0.020 0.095 0.235 0.328 0.244 0.075

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 1 2 3 4 5 6

P(x)

Relative Frequency

Households

NOTE: that the histogram is skewed left. The graph of a binomial distribution with p > .05 is skewed left, while the graph of a binomial distribution with p < .05 is skewed right. The graph of a binomial distribution with p = .05 is symmetric.

MEAN, VARIANCE AND STANDARD DEVIATION Although you can use the formulas

learned ifor mean, variance and standard deviation of a probability distribution, the properties of a binomial distribution enable you to use much simpler formulas. They are on the next slide.

POPULATION PARAMETERS OF A BINOMIAL DISTRIBUTION

Mean: = npVariance: 2 = npqStandard Deviation: = √npq

MEAN, VARIANCE AND STANDARD DEVIATION

Let's calculate the Mean, Variance and Standard Deviation for the Sports Bike inspections.

There are (relatively) simple formulas for them. They are alittle hard to prove, but they do work!

The mean, or "expected value", is:μ = n pFor the sports bikes:μ = 4 × 0.9 = 3.6So we would expect 3.6 bikes (out of 4) to pass the inspection. Makes sense really ... 0.9 chance for each bike times 4 bikes equals 3.6For the sports bikes: Variance: σ2 = 4 × 0.9 × 0.1 = 0.36Standard Deviation is: σ = √(0.36) = 0.6

FINDING MEAN, VARIANCE AND STANDARD DEVIATION

In Pittsburgh, 57% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy days during the month of June. What can you conclude?

Solution: There are 30 days in June. Using n=30, p = 0.57, and q = 0.43, you can find the mean variance and standard deviation as shown.

Mean: = np = 30(0.57) = 17.1Variance: 2 = npq = 30(0.57)(0.43) = 7.353Standard Deviation: = √npq = √7.353 ≈2.71

ADVANTAGESWell, as computers get more sophisticated, the advantage

is probably not as great as it once was. Still, to compute the sum of binomial probabilities for a huge number of trials, you will be asking a computer to add thousands of probabilities, each of which are very close to 0 (since they add up to one.) Every machine has limits of accuracy, and there is potential for round-off error. (And of course, once upon a time these would have had to have been added up by hand!) 

The normal approximation is quick and easy, and usually acurate enough for purposes of hypothesis testing.

LIMITATIONS

THE POISSON DISTRIBUTIONOVERVIEW When there is a large number

of trials, but a small probability of success, binomial calculation becomes impractical Example: Number of deaths

from horse kicks in the Army in different years

The mean number of successes from n trials is µ = np Example: 64 deaths in 20

years from thousands of soldiers

Simeon D. Poisson (1781-1840)

SUMMARY It deals with Bernoulli trails. We use binomial distribution when two

results are expected. It calculates the probability of an event by

finding a product of possible combinations with probability of a single combination.

THE BINOMIAL DISTRIBUTIONOVERVIEW