Stats 11 (Fall 2004) Lecture Note Instructor: Hongquan XuIntroduction to Statistical Methods for Business and EconomicsMidterm Exam 1 Review — Chapters 1, 2, 4 and 51. The histogram and b oxplot of the exam scores for 60 students are given below.Exam ScoreFrequency30 40 50 60 70 80 90 1000 2 4 6 8 10●30 40 50 60 70 80 90 100(a) What is the overall shap e of the distribution of the exam scores?(b) What proportion of students scored more than 90?(c) 25% of students scored more than .(d) What was the lowest score?(e) Based on the boxplot only, can you tell that the mean exam score was larger or less than themedian score? Explain why.12. A marketing research firm conducted a survey of companies in its state. They mailed a questionnaire to300 small companies, 300 medium-sized companies, and 300 large companies. The rate of nonresponseis important in deciding how reliable survey results are. Here are the data on response to this survey:ResponseSize of Company Yes No TotalSmall 175 125 300Medium 145 155 300Large 120 180 300(a) Draw a bar graph to compare the nonresponse rates for the three size categories. Label both axesclearly.(b) What is the overall nonresponse rate?(c) Given that a randomly selected company does not response to the survey, what is the probabilitythat this is a large company?(d) Are the variables “response” and “size of company” independent? You must fully explain youranswer for full credit.23. A basketball player has made about 70% of his free throws over several years. Suppose that his freethrows are independent trials with probability 0.7 of a success on each trial.(a) What is the probability that he makes none in 6 attempts?(b) What is the probability that he makes at least one in 6 attempts?4. A manufacturing process produces compact discs (CDs). It is known that 10% of the CD’s producedare defective. A simple test is used to determine whether a CD is defective, for which 98% of thedefective CDs are detected and 5% of the good CDs are declared defective.(a) What is the probability that a randomly selected CD will be declared defective by the test? Showall work.(b) Given that a randomly selected CD is declared defective by the te st, what is the probability thatthe CD is truly defective? Show all work.35. Let X be the number of persons living in an American household. Ignoring the few households withmore than seven inhabitants, the distribution of X is as follows:Inhabitants 1 2 3 4 5 6 7Probability 0.25 0.32 0.17 0.15 0.07 0.03 ?(a) What is the missing probability P (X = 7)?(b) What is P (2 < X ≤ 4)?(c) What is P (X 6= 1)?(d) What is P (X < 4|X 6= 1)?(e) What is the expected (mean) number of persons living in an American household?(f) What is the standard deviation of the number of persons living in an American household?Some important concepts for reviewDistinction b e tween experiments and observational study.The basic principles of design of expe riments.Sources of error in surveysRole of randomizationMeasures of centers and spread4Midterm Exam 1• Time: Monday, October 25, in class (10-10:50)• Material: Chapters 1, 2, 4 and 5 (including lectures, homework 1-3, labs 1-2)• It will be a closed book exam.• Bring your calculator, ruler, pen, pencil, eraser, etc.NOTE: The following formulas will be provided.Formulas.Sample Mean and Standard Deviationmean ¯x =1nPxi, standard deviation s =q1n−1P(xi− ¯x)2Probability RulesComplement Rule: P (Ac) = 1 − P (A).Addition Rule: P (A or B) = P (A) + P (B) − P (A and B).If A and B are disjoint (mutually exclusive), then P (A or B) = P (A) + P (B).Multiplication Rule: P (A and B) = P (A)P (B | A) = P (B)P (A | B)If A and B are indep endent then P (A and B) = P (A)P (B).Conditional Probability: P (A | B) =P (A and B)P (B)Partition Rule: If A1, A2, ..., Akform a partition,P (B) = P (B | A1)P (A1) + P (B | A2)P (A2) + ... + P (B | Ak)P (Ak)Bayes’ Rule: P (A | B) =P (B|A)P (A)P (B)=P (B|A)P (A)P (B|A)P (A)+P (B|Ac)P (Ac).Random Variablesmean E(X) =Pxf(x) and variance var (X) = E[(X − E(X))2] = E(X2) − [E(X)]2Rules for expectation: E[g(X)] =Pg(x)f (x) and E[g1(X) + g2(X)] = E[g1(X)] + E[g2(X)]Rules for means: E(a + cX) = a + cE(X)Rules for variances: var (a + cX) = c2var