Sampling WITHOUT replacement, Order IS important Number of Samples = 6

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1 : Different strategies sampling 2 out of numbers {1,2,3}: Sampling WITHOUT replacement, Order IS important Number of Samples = 6 (1,2) (1,3) (2,1) (2,3) (3,1) (3,2)

2 : Different strategies sampling 2 out of numbers {1,2,3}: Sampling WITHOUT replacement, Order IS NOT important Number of Samples = 3 (1,2) (1,3) (2,3) Usually, sample without replacement, thereby ensuring the sample is comprised of different individuals!

3 Permutations When sampling from the population WITHOUT replacement of the subjects and the order of sampled individuals IS important, the number of different arrangements, permutations of n subjects from a population of size N: where N! = [N factorial] =1*2*3* *(N-1)*N 0! = 1

4 : Permutations Locker combination (order is important) comprised of 3 distinct digits. Compute the number of different locker combination. N = n = P(N_n) = N!/(N-n)! =

5 : Permutations Locker combination (order is important) comprised of 3 distinct digits. Compute the number of different locker combination. N = 10 n = 3 P(N _ n) = N!/(N-n)! = 10!/(10-3)! = = (1*2* *7*8*9*10)/ (1*2* *7) = 720

6 Permutations Simple Properties General Formula

7 Combinations When sampling from the population WITHOUT replacement of the subjects and the order of sampled individuals IS NOT important, the number of different samples, combinations of n subjects from a population of size N:

8 Consider sampling 3 students from a class of 10 people. Compute the number of different samples. In this situation there will be no replacement, and the order of students in a sample is not important. N = n = C(N choose n) = N!/n!(N-n)! =

9 Consider sampling 3 students from a class of 10 people. Compute the number of different samples. N =10 n = 3 C(N choose n) = N!/n!(N-n)! =10!/(3!7!) = =8*9*10/(2*3) = 120

10 Combinations Simple Properties General Formula

11 You have N= close friends that you would like to invite to the movie. In how many ways can you invite 2 of them? Is order important? What do you have to compute: Permutation or Combination?

12 You have N= close friends that you would like to invite to the movie. In how many ways can you invite 2 of them? Order is Not important Compute Combination: C(N_2) = N*(N-1)/2

13 Binomial Random Variables There are n trials (performances of the binomial experiment), where n is determined, not random. Each trial results in only 1 outcome: success or failure. The probability of a success on each trial is constant, denoted p, with 0 p 1. The probability of a failure q=1-p, with 0 q 1. The trials are independent.

14 A coin is tossed 3 times (n=3). Each time there are two possible outcomes: tail (success) and head (failure), with constant probabilities p and q (q=1-p). Define random variable X as the number of tails in 3 independent trials. X is a discrete random variable that takes the values 0, 1, 2, 3. X ~ Binomial(n=3, p =0.5)

15 : Are Conditions satisfied for Binomial Model? Observe the gender of the next 50 children born at a local hospital (X= number of girls.) Yes No A ten-question quiz has 5 True-False questions and 5 multiple-choice questions, each with 4 possible choices. A student randomly picks an answer for event question. X= number of correct answers. Yes No

16 X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X P(X=3) =?

17 X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X P(X=3) = p*p*p = p^3 P(X=0) = q^3

18 X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X P(X=2) =?

19 X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X P(X=2) = p*p*q + p*q*p + q*p*p = 3p^2*q Recall C(3_2) = 3!/(2!(3-2)!) = 3

20 X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X P(X=1) =?

21 X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p Trial 1 T T T H T H H H Trial 2 T T H T H T H H Trial 3 T H T T H H T H X P(X=1) = p*q*q + q*p*q + q*q*p = 3p*q^2 Recall C(3_1) = 3!/(1!(3-1)!) = 3

22 X as the number of tails in 3 independent trials X~ Binomial(n=3, p =0.5) P( tail ) = p, P( head ) = q = 1-p X=k P(X=k) q^3 3pq^2 3p^2q p^3

23 Binomial Distribution X~ Binomial (n, p) Probability of exactly k successes in n trials: where Mean (Expected value) of X is Standard Deviation of X is

24 A restaurant manager has calculated, based on years of experience, that about 20% of the patrons order the chef s steak special each Friday evening. Consider taking a random sample of patrons and let X be the number of sampled patrons who will order the chef s steak special on a Friday evening.

25 What is the exact distribution of X for a random sample of 10 patrons? (Provide all features).

26 What is the exact distribution of X for a random sample of 10 patrons? (Provide all features). X has a Binomial distribution with sample size n = 10 and probability of success p = 0.2.

27 What is the probability that exactly one patron in the sample of 10 will order the chef s steak special on a Friday evening? P(X=1) =?

28 What is the probability that exactly one patron in the sample of 10 will order the chef s steak special on a Friday evening? n=10, k=1, p=0.2 C(n_k) = n!/(k!(n-k)!) = 10!/(1!(10-1)!) = 10 P(X=1) =10*(0.2)^1*(1-0.2)^9 =

29 What is the probability that at least one patron in the sample of 10 will order the chef s steak special on a Friday evening? P(X>=1) =?

30 P( X Example What is the probability that at least one patron in the sample of 10 will order the chef s steak special on a Friday evening? P(X=>1) = P(X=1) + P(X=2) + +P(X=10) Hint: Key here is to use the complement rule. P(at least one) = 1 P(none). 1) 1 P( X 0) (0.2) 0 (1 0.2)

31 How many patrons in the sample of 10 you expect to order the chef s steak special on a Friday evening? For Binomial distribution E(X)=μ=np= 10*0.2=2

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