Revision Applied Statistics PDF
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Canadian University Dubai
Canadian University Dubai
Dr. Noura Mansoura
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This document is a past paper for applied statistics, containing examples and problems related to probability distributions, conditional probability, and other statistical concepts covered in a statistics course. This document may be useful in the preparation for a future exam in applied statistics.
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1 Revision For the Final Exam Dr. Noura Mansoura From Chapter 5 to chapter 12: Example 1: If a family plans to have three children, draw a tree diagram and list the sample space showing all possible arrangements of the...
1 Revision For the Final Exam Dr. Noura Mansoura From Chapter 5 to chapter 12: Example 1: If a family plans to have three children, draw a tree diagram and list the sample space showing all possible arrangements of the children's genders with respect to their ages. Required: I. Determine the elements of the following probabilities: A. All children are of the same gender. B. At most, one child is a boy C. There is only one boy, given the middle child is a girl II. Then find P(A), P(B), P(A∩B), P(A∪B) and P(C) Sample Space S={BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG} (A) All children are of the same gender: {BBB, GGG} (B) At most, one child is a boy At most, one boy means there can be either 0 or 1 boy: {BGG,GBG, GGB,GGG}. 1 2 (C) There is only one boy, given the middle child is a girl The middle child is a girl, so consider only outcomes where the second child is G: {BGG,GBG,GGB,GGG} , Among these, outcomes of exactly one boy: {BGG,GGB} is the answer 1. P(A)=0.25 2. P(B)=0.5 3. P(A∩B)=0.125 4. P(A∪B)=0.625 5. P(C)=0.5 ( you could solve it: Let D: There is only one boy and E: The middle child is a girl So, D={bgg,gbg,ggb} P(D)=3/8 And E = B={bgb,bgg, ggb, ggg} P(D)=4/8 D∩E = {bgg, ggb } P(D∩E)=2/8 = (2/8)/(4/8)=2/4= ½ =0.5 Example 2: The following are the data on the gender and marital status of 100 customers of a company. Male Female In total Single 20 30 Married 10 40 50 In total 30 70 100 a. What percentage of male customers? (30/100)*100=30% b. What is the probability of finding a married male customer? 10/100 2 3 c. What is the probability that a randomly selected customer is a male or married? P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) P(Male U Married)=30/100+50/100-10/100=70/100 d. If a customer is male, what is the probability that he is married? what is the probability that the selected customer is married, if (given that) a customer is male P( A ∩ B) P( A B) =. P( B) 10/100 P(Married|Male)= =10/30 30/100 e. Are gender and marital status mutually exclusive? Explain. None of the cells is zero, therefore the intersection is not zero, they will never be mutually exclusive f. Is marital status independent of gender? Explain. ↔ P ( A ∩ B ) = P ( A) P ( B ) 10/100≠30/100*50/100, therefore they are not equal, so they are not independent Example 3: A factory produces items, and the probability that an item is defective is 5%. If a random sample of 12 items is selected, Find the probability that: a. Exactly 2 items are defective. b. Fewer than 3 items are defective. c. 4 or more items are defective. d. between 1 and 3, inclusive. e. Calculate the mean and standard deviation This one is binomial; we have p= 0.05, and n is fixed 12. a. P(X=2)=66*0.0025*0.5987≈0.0988 b. P(X=4)= 1- P(X