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The Binomial Probability Distribution. ● A binomial experiment has the following structure  The first test is performed … the result is either a success.

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Presentation on theme: "The Binomial Probability Distribution. ● A binomial experiment has the following structure  The first test is performed … the result is either a success."— Presentation transcript:

1 The Binomial Probability Distribution

2 ● A binomial experiment has the following structure  The first test is performed … the result is either a success or a failure  The second test is performed … the result is either a success or a failure. This result is independent of the first and the chance of success is the same  A third test is performed … the result is either a success or a failure. The result is independent of the first two and the chance of success is the same

3  A card is drawn from a deck. A “success” is for that card to be a heart … a “failure” is for any other suit  The card is then put back into the deck  A second card is drawn from the deck with the same definition of success.  The second card is put back into the deck  We continue for 10 cards

4 ● A binomial experiment is an experiment with the following characteristics  The experiment is performed a fixed number of times, each time called a trial  The trials are independent  Each trial has two possible outcomes, usually called a success and a failure  The probability of success is the same for every trial

5  Notation used for binomial distributions  The number of trials is represented by n  The probability of a success is represented by p  The total number of successes in n trials is represented by X  Because there cannot be a negative number of successes, and because there cannot be more than n successes (out of n attempts) 0 ≤ X ≤ n

6 ● In our card drawing example  Each trial is the experiment of drawing one card  The experiment is performed 10 times, so n = 10  The trials are independent because the drawn card is put back into the deck  Each trial has two possible outcomes, a “success” of drawing a heart and a “failure” of drawing anything else  The probability of success is 0.25, the same for every trial, so p = 0.25  X, the number of successes, is between 0 and 10

7  The word “success” does not mean that this is a good outcome or that we want this to be the outcome  A “success” in our card drawing experiment is to draw a heart  If we are counting hearts, then this is the outcome that we are measuring  There is no good or bad meaning to “success”

8  We would like to calculate the probabilities of X, i.e. P(0), P(1), P(2), …, P(n)  Do a simpler example first  For n = 3 trials  With p =.4 probability of success  Calculate P(2), the probability of 2 successes

9 ● For 3 trials, the possible ways of getting exactly 2 successes are  S S F  S F S  F S S ● The probabilities for each (using the multiplication rule) are  0.4 0.4 0.6 = 0.096  0.4 0.6 0.4 = 0.096  0.6 0.4 0.4 = 0.096

10 ● The total probability is P(2) = 0.096 + 0.096 + 0.096 = 0.288 ● But there is a pattern ● The probability for each case is 0.4 2 0.6 1

11 ● There are 3 cases  S S F could represent choosing a combination of 2 out of 3 … choosing the first and the second  S F S could represent choosing a second combination of 2 out of 3 … choosing the first and the third  F S S could represent choosing a third combination of 2 out of 3 ● These are the 3 = 3 C 2 ways to choose 2 out of 3

12 ● Thus the total probability P(2) =.096 +.096 +.096 =.288 can also be written as P(2) = 3 C 2.4 2.6 1 ● In other words, the probability is  The number of ways of choosing 2 out of 3, times  The probability of 2 successes, times  The probability of 1 failure

13 ● The general formula for the binomial probabilities is just this ● For P(x), the probability of x successes, the probability is  The number of ways of choosing x out of n, times  The probability of x successes, times  The probability of n-x failures ● This formula is P(x) = n C x p x (1 – p) n-x

14  A student guesses at random on a multiple choice quiz  There are n = 10 questions in total  There are 5 choices per question so that the probability of success p = 1/5 =.2  What is the probability that the student gets 6 questions correct?

15  This is a binomial experiment  There are a finite number n = 10 of trials  Each trial has two outcomes (a correct guess and an incorrect guess)  The probability of success is independent from trial to trial (every one is a random guess)  The probability of success p =.2 is the same for each trial

16 ● The probability of 6 correct guesses is P(x)= n C x p x (1 – p) n-x = 10 C 6.2 6.8 4 = 210.000064.4096 =.005505 ● This is less than a 1% chance ● In fact, the chance of getting 6 or more correct (i.e. a passing score) is also less than 1%

17  We would like to find the mean of a binomial distribution  Example  There are 10 questions  The probability of success is.20 on each one  Then the expected number of successes would be 10.20 = 2  The general formula μ X = n p

18  We would like to find the standard deviation and variance of a binomial distribution  This calculation is more difficult  The standard deviation is σ X = √ n p (1 – p) and the variance is σ X 2 = n p (1 – p)

19 ● For our random guessing on a quiz problem  n = 10  p =.2  x = 6 ● Therefore  The mean is np = 10.2 = 2  The variance is np(1-p) = 10.2.8 = 1.6  The standard deviation is √1.6 = 1.26 ● Remember the empirical rule? A passing grade of 6 is 10 standard deviations from the mean …

20 ● With the formula for the binomial probabilities P(x), we can construct histograms for the binomial distribution ● There are three different shapes for these histograms  When p <.5, the histogram is skewed right  When p =.5, the histogram is symmetric  When p >.5, the histogram is skewed left

21  For n = 10 and p =.2 (skewed right)  Mean = 2  Standard deviation =.4

22  For n = 10 and p =.8 (skewed left)  Mean = 8  Standard deviation =.4

23  For n = 10 and p =.5 (symmetric)  Mean = 5  Standard deviation =.5

24  Despite binomial distributions being skewed, the histograms appear more and more bell shaped as n gets larger  This will be important!

25  Suppose 70% of all Americans have cable TV.  In a Random sample of 15 households, what is the probability that exactly 10 households have cable?  Binompdf(n, p, x)  Binompdf(15,.7, 10)  Use pdf when exactly is used!!!  P(x = 10)=

26  Suppose 70% of all Americans have cable TV.  In a Random sample of 15 households, what is the probability that at least 13 households have cable? 1- Binomcdf(n, p, x)  P( x 13) = P(13) or P(14) or P(15)

27  Suppose 70% of all Americans have cable TV.  In a Random sample of 15 households, what is the probability that fewer than 13 households have cable?  Notice the 12….NOT 13!!  P(x 12)= P(1)+P(2)….P(12)

28  According to flightstats.com, American Airlines flight 1247 from Orlando to Los Angeles is on time 65% of the time. Suppose fifteen flights are randomly selected, and the number of on-time flights is recorded. a) Explain why this is a binomial experiment b) Find the probability that exactly 10 flights are on time. c) Find the probability that at least 10 flights are on time. d) Find the probability that fewer that 10 flights are on time. e) Find the probability that between 7 and 10 flights, inclusive, are on time.

29 According to the Uniform Crime Report, 2006, nationwide, 61% of murders committed in 2006 were cleared by arrest or exceptional means. a) For 250 randomly selected murders committed in 2006, compute the mean and standard deviation of the random variable x, the number of murders cleared by arrest or exceptional means. b) Interpret the mean c) Of the 250 randomly selected murders, find the interval that would be considered “unusual” for the number of murders that was cleared.


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