1. BINOMIAL AND NORMAL DISTRIBUTIONS BINOMIAL DISTRIBUTION “Bernoulli trials” – experiments satisfying 3 conditions: 1. Experiment has only 2 possible outcomes: Success, S and Failure, F. 2. The probability of S is fixed (does not change) from trial to trial. P(S)=p, 0<p<1, P(F)= 1- P(S)=1-p. 3. n independent trials of the experiment are performed. Let X= # of S in n Bernoulli trials. X has Binomial distribution with number of trials n and probability of success p. X~Bin(n, p) X is a discrete r.v. with 2 parameters: n and p. X counts number of S in n Bernoulli trials (Binomial type experiment).

2. EXAMPLES 1. Toss a coin 10 times. Record the number of H. Is this Binomial type experiment? 2. Toss a coin until you get a T. Record the number of tosses. Is this Binomial type experiment? 3. Toss a die 20 times. Record the number of “5”. Is this Binomial type experiment? 4. Toss a die 20 times. Record the number of times an “even” face comes up. Is this Binomial type experiment? 5. Observe weather in Seattle on 100 days. Record if it rains on a given day. Is this Binomial type experiment? 6. Select 5 cards from a deck of 52. Record the number of Aces. Is this Binomial type experiment? 7. Select 5 cards from a deck of 52. For each card record if it is an Ace. Is this Binomial type experiment?

3. Probability distribution of Bin(n,p) r.v. n = # of trials,. An outcome of the experiment with k successes, (0≤k≤n) and n-k failures, for example: has probability P(k successes out of n trials) =# ways to place k S among n trials x p k (1-p) n-k # ways to place k S among n trials = = and Finally, P( k successes out of n trials ) =

4. Probability distribution of a Bin(n,p) r.v. X~Bin(n, p), n = number of trials, p = probability of S, 0 < p < 1. Values of X: 0, 1, …, n. P(k successes out of n trials) = P(X= k) = Example: NOTE: Table B in the Appendix lists Binomial probabilities for n=2, 3, …, 20 and p=0.1, 0.2, 0.25, …, 0.9.

5. EXAMPLE What is the probability that in a family of 5 children 2 are girls? What is the probability of having all girls? Solution. Trial/experiment: parents have a child Girl or Boy S F Our family- 5 children i.e. 5 trials, p=P(S)=P( girl )=0.5. X= # girls among 5 children; X~Bin(5, 0.5). P(X = 2) = Probability of having all girls? P(X=5)=

6. EXAMPLE A commuter plane has 10 seats. The airline books 12 people on the flight. Suppose the chance of a person who makes a reservation of actually showing up is 0.8. Find the probability that someone is bumped and the probability that at least one seat is empty. Solution. Trial/experiment: A person with reservation decides: Show up OR Do not show up for the flight S F Total # of people with reservations =12. Total # of trials= 12. P(S) = 0.8 X= # people who show up for the flight; X~Bin(12, 0.8). I used Table B for binomial probabilities. P(someone is bumped) = P(more than 10 people show up)= P(11 or 12 people show up)=P(X=11 or X=12) = P(X=11) + P(X=12) = = 0.2062 + 0.0687 = 0.2749. P(at least one seat is empty)= P(at most 9 people showed up)= 1- P(X=10or X=11 or X=12) = 1- ( 0.2062+0.0687+0.2835) = 0.4416.

7. Mean and variance of a binomial random variable If X~Bin(n, p) random variable, then the mean of X, μ x = EX=np And the standard deviation of X, NOTES: 1. Variance of X is σ 2 =np(1-p). 2. The mean of a binomial r.v. (mean number of successes) is the number of trials x the probability of success.

8. EXAMPLE 1. Fair coin was tossed 3 times. Let X =# of heads in the 3 tosses. What is the mean and standard deviation of X? Solution. X~Bin(3, 0.5). Mean of X μ x = EX=np=3(0.5)=1.5. Standard deviation of X is 2. The overbooking airline example. What is the mean and standard deviation of the number of passengers that show up for the flight? Solution. X~Bin(12, 0.8). Mean of Xμ x = EX=np=12(0.8)=9.6. Standard deviation of X is