Week 3 tests.docx - Question 1 of 20 0.0 1.0 Points If...

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Question 1 of 20 0.0/ 1.0 Points If random variable X has a binomial distribution with n =10 and P(success) =p =0.2, find the probability that X is less than 6. (That is, find P(X<6) Answer: (round to 4 decimal places)
Question 2 of 20 0.0/ 1.0 Points Approximately 8% of all people have blue eyes. Out of a random sample of 20 people, what is
Question 3 of 20 1.0/ 1.0 Points It is known that 40% of adult workers have a high school diploma. If a random sample of 10 adult workers is selected, what is the expected number of adult workers with a high school
Question 4 of 20 1.0/ 1.0 Points Approximately 10% of all people are left-handed. If 200 people are randomly selected, what is the expected number of left-handed people? Round to the whole number. Do not use decimals.
Answer Key:20 Feedback: 200*.10 Question 5 of 20 0.0/ 1.0 Points If random variable X has a binomial distribution with n=15 and P(success) =p= 0.6, find the
Part 2 of 6 - Contingency Table Knowledge Check Practice 1.0/ 1.0 Points Question 6 of 20 1.0/ 1.0 Points The table of data obtained from shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected. NAME Single Double Triple Home Run TOTAL HITS Babe Ruth 1,517 506 136 714 2,873 Jackie Robinson 1,054 273 54 137 1,518 Ty Cobb 3,603 174 295 114 4,189 Hank Aaron 2,294 624 98 755 3,771 TOTALS 8,468 1,577 583 1,720 12,351 Find P (hit was a Home Run|The hit was made by Ty Cobb ).
0.027 C. 0.009 D. 0.973 Answer Key:B Feedback:114/4189 Part 3 of 6 - Counting Principle Knowledge Check Practice 0.0/ 3.0 Points Question 7 of 20 0.0/ 1.0 Points The number of M&M’s for each color found in a case were recorded in the table below. Blue = 481, Brown = 371, Green = 483, Orange = 544, Red = 372, Yellow = 369 Total = 2620
Question 8 of 20 0.0/ 1.0 Points Find the probability of rolling a sum of two dice that is a 7 or a 12. Round answer to 4 decimal places. Answer:
Feedback: Dice outcomes 12 3 4 5 6 1 (1, 1) (1, 2) (1, 3)(1, 4)(1, 5)(1, 6) 2 (2, 1) (2, 2) (2, 3)(2, 4)(2, 5)(2, 6) 3 (3, 1) (3, 2) (3, 3)(3, 4)(3, 5)(3, 6) 4 (4, 1) (4, 2) (4, 3)(4, 4)(4, 5)(4, 6) 5 (5, 1) (5, 2) (5, 3)(5, 4)(5, 5)(5, 6) 6 (6, 1) (6, 2) (6, 3)(6, 4)(6, 5)(6, 6) Only 7 ways to sum to exactly 7 or exactly 12 7/36

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