Lecture 8. The Binomial Distribution. Binomial Distribution. Binomial Distribution. Probability Distributions: Normal and Binomial

Transcription

1 Lecture 8 The Binomial Distribution Probability Distributions: Normal and Binomial 1 2 Binomial Distribution >A binomial experiment possesses the following properties. The experiment consists of a fixed number (n) of trials The result of each trial can be classified into one of two categories: success or failure The probability (π) of a success remains constant for each trial Each trial of the experiment is independent of the other trials 3 Binomial Distribution APPROACHES TO SOLVING: >FORMULA >TABLES >APPROXIMATION USING THE NORMAL DISTRIBUTION - later 4

2 Formula Approach Notation n! x) ( ) p ( ) ( ) x q (n P x = x! n x! Where: P(x) = the probability of x successes in n trials n = the number of trials p = the probability of success on any one trial q = (1 - p) C n x = x! n = x x! n! ( n x)! n! ( ) n x! > You may be familiar with one of the notations at the left > It is the number of ways of choosing x objects from a total of x objects > The notation x! means the product of all the numbers from 1 to x 5 6 >80% of transactions handled by CSRs could normally be handled by ATMs or over the phone. During a slow period in the afternoon, 5 customers were helped by CSRs. >What is the probability that all could have been served without CSRs? >What is the probability that 3 needed CSR intervention? Formula Solution > p = 0.80, i.e. Prob. of not needing CSR >n = 5 > x = 5, x = 5-3 = 2 p(5) = C ( 0.8) ( 0.2) p(2) = C 5 2( 0.8) ( 0.2) 5! 5 0 5! 2 3 = ( 0.8) ( 0.2) = ( 0.8) ( 0.2) 5! 0! 2! 3! = ( 10)( 0.64)( 0.008) = = % = = 5.12% 7 8

3 Binomial Tables > The tables in text Appendix B (pages ) give binomial probabilities for n = 1 through 15, 20 and 25 trials and a range of probabilities > Below, is a table for n=5 n = 5 π x n = 5 π x P(X=2) P(X=5) 9 10 Binomial Expected Value, Variance & Standard Deviation >Expected Value: E(x) = µ = np >Variance: Var(x) = σ 2 = npq >Standard Deviation SD(x) = σ = npq Binomial Recognition >Each result is binary, i.e. a success or failure, good or bad, yes or no DISCRETE NO FRACTIONAL VALUES >Maximum = n, Minimum =0 >The probability of any result remains constant unlike lotto, once a number is taken, it cannot be chosen again >Each result is independent 11 12

4 Binomial Exercise >The probability of a statistics student doing their assignment more than 24 hours before it is due is 25%. If ten students are chosen at random the day before it is due, what is the probability that none have done it? 2 have done it? more than 3 have done it? less than 5 have done it? 13 x Tables Approach P(X=0) = P(X=2) = P(X>3) = P(X<5) = P(X = 0) = P(X = 2) = Using the Formula 10! 0! ( )( ) ( ) ( 0.75 ) 10! 10! 2! 2 = ( )( ) ( 0.25) ( 0.75) = 8! > For P(X>3) and P(X<5) we would need to use the formula many times, which is generally not practical 8 10 >Four out of five dentists recommend Crest. If you survey 10 dentists, what is the probability that: More than 4 out of 5 recommend Crest? Less than 4 out of 5 recommend Crest? Exactly 4 out of 5 recommend Crest? 15 16

5 x Table Less than 4 out of Exactly 4 out of More than 4 out of >A student believes that she will be accepted to 80% of the universities to which she applies. What is the probability not being accepted to any university if she applies to 4 universities? >P(X=0) = (0.2) 4 = Normal Distribution Normal Distribution >We have seen the normal distribution several times before >Often referred to as the Bell Curve >Occurs often in nature 19 20

6 Normal Distribution Normal Properties Probability Density >All Normal Distributions are simply scaled versions of each other >Changing the mean shifts the distribution right or left >Changing the standard deviation makes it flatter or narrower Z Value- Standard Deviations from Mean Standard Normal > All normals can be converted to the standard normal (mean, 0; standard deviation, 1) by subtracting the mean from each value and dividing by the standard deviation Z = X - µ σ x x Standard Normal >This transformation results in Z being distributed with a mean of 0 and a standard deviation of 1 >Z simply measures a distance: the number of standard deviations above or below

7 Standard Normal >Thus, for any value or observation, its Z-value lets us know how many standard deviations it is above or below zero >This is the simplicity of the normal distribution: All normally distributed data can be converted to a standard normal Standard Normal >This means that there is only one table that we need to reference and learn >For any normally distributed variable, the distribution of its probabilities are identical to all other normally distributed data or variable - both can be converted into the standard normal Understanding The Normal >The Normal Distribution is SYMMETRICAL about its mean half above & half below >The normal is infinite in both directions There are very small probabilities of extreme values Understanding The Normal >The probability of a value falling within a specified number of standard deviations is given by the well known EMPIRICAL RULE that we looked at in Lecture

8 Empirical Rule Normal Distribution Number of Std. Dev. Proportion of Distribution Approximately Exactly ±1 2/3 68.4% ±2 95% 19 out of % ±3 Substantially all 99.7% Probability Density 0.15% Tail 2.5% Tail 2/3 95% 99.7% 2.5% Tail 0.15% Tail Z Value- Standard Deviations from Mean 30 Normal Distribution APPROACHES TO SOLVING: >TABLES >COMPUTER - not covered in this course Normal Table >The normal table is in text Appendix D on page 743 >Only half the table is given, since the Normal Distribution is symmetrical >The values given is the P(0< z ), the area between Z=0 and Z=z 31 32

9 Normal Table (Part) Second Decimal Place z P(Z< 0.98) In Class Exercise >A statistics midterm had an average of 75 and a standard deviation of 10. >What is the probability that a student chosen at random received: At least B+? Less than C+? B+? B or B+? C or C+? Question Normal Standardized At least B+ P(X>75) P(Z>0.0) Less than C+ P(X<65) P(Z<-1.0) B+ P(75<X<80) P(0<Z<0.5) B or B+ P(70<X<80) P(-0.5<Z<0.5) C or C+ P(60<X<70) P(-1.5<Z<-0.5) Question Standardized From Table At least B+ P(Z>0.0) = 0.5 Less than C+ P(Z<-1.0) =P(Z>1.0) = B+ P(0<Z<0.5) B or B+ P(-0.5<Z<0.5) =P(Z<0.5)x x2 = C or C+ P(-1.5<Z<-0.5) =P(0.5<Z<1.5) =

10 YOU LEARN STATISTICS BY DOING STATISTICS 37