Q #1

Isabel Briggs Myers was a pioneer in the study of personality types. The personality types are broadly defined according to four main preferences. Do married couples choose similar or different personality types in their mates? The following data give an indication. Similarities and Differences in a Random Sample of 375 Married Couples Number of Similar Preferences Number of Married Couples All four 29 Three 135 Two 124 One 66 None Suppose that a married couple is selected at random. (a) Use the data to estimate the probability that they will have 0, 1, 2, 3, or 4 personality preferences in common. (Enter your answers to 2 decimal places.) 0 1 2 ................. 3 .......... 4 l0.10 lX ions l.' l0.32 lxgoss x 0.03 'p (b) Do the probabilities add up to 1? Why should they? 0 Yes, because they do not cover the entire sample space. 0 No, because they do not cover the entire sample space. (D Yes, because they cover the entire sample space. 0 No, because they cover the entire sample space. What is the sample space in this problem? O 0, 1, 2, 3 personality preferences in common O 1, 2, 3, 4 personality preferences in common 0 0, 1, 2, 3, 4, 5 personality preferences in common 0, 1, 2, 3, 4 personality preferences in common (a) If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely? O 0, 1, 2, 3, 4, 5, 6; equally likely 1, 2, 3, 4, 5, 6; equally likely O 1, 2, 3, 4, 5, 6; not equally likely O 0, 1, 2, 3, 4, 5, 6; not equally likely V (b) Assign probabilities to the outcomes of the sample space of part (a). (Enter your answers as fractions.) Outcome Probability 1 1/6 4 2 1/3 4 3 1/6 4 4 1/3 4 5 1/6 4 5 1/5 4 Do the probabilities add up to 1? Should they add up to 1? Explain. 0 Yes, but they should not because these values do not cover the entire sample space. 6) Yes, because these values cover the entire sample space. 0 No, because these values do not cover the entire sample space. 0 No, but they should because these values cover the entire sample space. (c) What is the probability of getting a number less than 4 on a single throw? (Enter your answer as a fraction.) 66.67 x (d) What is the probability of getting 3 or 4 on a single throw? (Enter your answer as a fraction.) Z Need Help? When do creative people get their best ideas? USA Today did a survey of 966 inventors (who hold U.S. patents) and obtained the following information. Time of Day When Best Ideas Occur Time Number of Inventors 6 A.M.-12 noon 278 12 noon6 RM. 137 6 P.M.12 midnight 314 12 midni ht6 A.M. 237 (a) Assuming that the time interval includes the left limit and all the times up to but not including the right limit, estimate the probability that an inventor has a best idea during each time interval: from 6 A.M. to 12 noon, from 12 noon to 6 P.M., from 6 RM. to 12 midnight, from 12 midnight to 6 A.M. (Enter your answers to 3 decimal places.) 6AM-12PM 12PM-6PM 6PM-12AM 12AM-6AM (b) Do the probabilities add up to 1? Why should they? O No, because they do not cover the entire sample space. (9 Yes, because they cover the entire sample space. 0 Yes, because they do not cover the entire sample space. 0 No, because they cover the entire sample space. VI What is the sample space in this problem? 0 12AM-12PM O 6AM-6PM O 6AM12AM C) the entire day A botanist has developed a new hybrid cotton plant that can withstand insects better than other cotton plants. However, there is some concern about the germination of seeds from the new plant. To estimate the probability that a seed from the new plant will germinate, a random sample of 3000 seeds was planted in warm, moist soil. Of these seeds, 2260 germinated. (a) Use relative frequencies to estimate the probability that a seed will germinate. What is your estimate? (Enter your answer to 3 decimal places.) : (b) Use relative frequencies to estimate the probability that a seed will not germinate. What is your estimate? (Enter your answer to 3 decimal places.) : (c) Either a seed germinates or it does not. What is the sample space in this problem? O germinate germinate or not germinate O not germinate Do the probabilities assigned to the sample space add up to 1? Should they add up to 1? Explain. Yes, because they cover the entire sample space. 0 Yes, because they do not cover the entire sample space. 0 No, because they cover the entire sample space. 0 No, because they do not cover the entire sample space. J (d) Are the outcomes in the sample space of part (c) equally likely? 0 yes Betting odds are usually stated against the event happening (against winning). The odds against event W are the ratio M = M. P(W) P0\") In horse racing, the betting odds are based on the probability that the horse does not win. a a+b' (a) Show that if we are given the odds against an event W as a:b, the probability of not W is given by P(W':) = P(W) = E P(not W) Pgnot W _ 4 PM) ' E Pgnot W) a b[P(not m] = E {1 P(not W)] b[P(not W)] + a[P(not W)] = E (a + b)[P(not m] = E :l PM\" W) = E (b) [n a recent Kentucky Derby, the betting odds for the favorite horse were 8 to 4. Use these odds to compute the probability that the favorite horse would lose the race. What is the probability that the favorite horse would win the race? (Round your answers to two decimal places.) (c) In the same race, the betting odds for a second horse were 6 to 1. Use these odds to estimate the probability that this horse would lose the race. What is the probability that this horse would win the race? (Round your answers to two decimal places.) (d) One of the horses was a long shot, with betting odds of 25 to 1. Use these odds to estimate the probability that the long shot would lose the race. What is the probability the horse would win the race? (Round your answers to two decimal places.) Pause =: mm =: Need Help? _ M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the distribution of colors for plain M&M candies is as follows. Color Purple Yellow Red Orange Green Blue Brown Percentage 21% 17% 20% 6% 10% 9% 17% Suppose you have a large bag of plain M&M candies and you choose one candy at random. (a) Find P(green candy or blue candy). : Are these outcomes mutually exclusive? Why? O No. Choosing a green and blue M&M is possible. O Yes. Choosing a green and blue M&M is possible. O No. Choosing a green and blue M&M is not possible. 6) Yes. Choosing a green and blue M&M is not possible. I (b) Find P(ye|low candy or red candy). : Are these outcomes mutually exclusive? Why? O No. Choosing a yellow and red M&M is not possible. O Yes. Choosing a yellow and red M&M is possible. (9 Yes. Choosing a yellow and red M&M is not possible. O No. Choosing a yellow and red M&M is possible. J \f9. [0.25/1 Points] DETAILS PREVIOUS ANSWERS BBUNDERSTAT12 4.2.018. MY NOTES ASK YOUR TEACHER You roll two fair dice, one green and one red. (a) Are the outcomes on the dice independent? Yes 0 No J (b) Find P(1 on green die and 2 on red die). (Enter your answer as a fraction.) : (c) Find P(2 on green die and 1 on red die). (Enter your answer as a fraction.) (d) Find P((1 on green die and 2 on red die) or (2 on green die and 1 on red die)). (Enter your answer as a fraction.) Need Help? @ A recent study gave the information shown in the table about ages of children receiving toys. The percentages represent all toys sold. A e ears Percenta eof To 5 2 and under 17% 3-5 19% 69 28% 10-12 11% 13 and over 25% What is the probability that a toy is purchased for someone in the following age ranges? (a) 6 years old or older (b) 12 years old or younger (c) between 6 and 12 years old (d) between 3 and 9 years old A child between 10 and 12 years old looks at this probability distribution and asks, "Why are people more likely to buy toys for kids older than I am (13 and over) than for kids in my age group (1012)?" How would you respond? 0 The 13-and-older category may include children up to 17 or 18 years old. This is a larger category. 0 The 13andolder category may include children up to 17 or 18 years old. This is a smaller category. Need Help? 13. [-11 Points] DETAILS BBUNDERSTAT12 4.2.026. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Based on data from a statistical abstract, only about 20% of senior citizens (65 years old or older) get the flu each year. However, about 29% of the people under 65 years old get the u each year. In the general population, there are 14% senior citizens (65 years old or older). (Round your answers to three decimal places.) (a) What is the probability that a person selected at random from the general population is senior citizen who will get the u this season? S (b) What is the probability that a person selected at random from the general population is a person under age 65 who will get the u this year? S (c) Repeat parts (a) and (b) for a community that has 92% senior citizens. (a): (b): (d) Repeat parts (a) and (b) for a community that has 47% senior citizens. (a): (b): Need Help? 14. [-11 Points] TAILS BBUNDERSTAT12 4.2.028. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A researcher says that there is a 72% chance a polygraph test (lie detector test) will catch a person who is, in fact, lying. Furthermore, there is approximately a 9% chance that the polygraph will falsely accuse someone of lying. (a) If the polygraph indicated that 30% of the questions were answered with lies, what would you estimate for the actual percentage of lies in the answers? Hint: Let B = event detector indicates a lie. We are given P(B) = 0.3. Let A = event person is lying, so AC = event person is not lying. Then P(B) = PM and B) + P(AC and B) P(B) = P(A)P(B | A) + P(AC)P(B | AC) Replacing P(A':) by 1 P(A) gives P03) = PM) ' P(B IA) + [1 - P(A)] ~ P03 | AC) Substitute known values for P(B), P(B IA), and P(B l AC) into the last equation and solve for P(A). (Round your answer to two decimal places.) m = E (b) If the polygraph indicated that 70% of the questions were answered with lies, what would you estimate for the actual percentage of lies? (Round your answer to one decimal place.) Need Help? In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept: Sale No Sale Row Total Aggressive 272 308 580 Passive 491 89 580 Column Total 763 397 1160 Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: A = aggressive approach, Pa = passive approach, S = sale, N = no sale. 50, P(A) is the probability that an aggressive approach was used, and so on. (a) Compute P(S), P(S IA), and P(S | Pa). (Enter your answers as fractions.) PtS) =: MS I A) =: P6 I Pa) =: (b) Are the events 5 = sale and Pa = passive approach independent? Explain. O Yes. P(S) = P(S | Pa). O Yes. The two events can occur together: O No. P(S) x P(S | Pa). O No. The two events cannot occur together. (c) Compute P(A and S) and P(Pa and 5). (Enter your answers as fractions.) Pm s> =: P(Pa s> =: (d) Compute P(N) and P(N | A). (Enter your answers as fractions.) P(N) =: P(N l A) =: (b) Are the events 5 = sale and Pa = passive approach independent? Explain. 0 Yes. P(S) = P(S | Pa). 0 Yes. The two events can occur together. 0 No. P(S) a: P(S | Pa). O No. The two events cannot occur together. (c) Compute PM and S) and P(Pa and 5). (Enter your answers as fractions.) Pm s> =: m 5) =: (d) Compute P(N) and P(N | A). (Enter your answers as fractions.) PM!) =: PM I A) =|:I (e) Are the events N = no sale and A aggressive approach independent? Explain. O No. The two evens cannot occur together. 0 Yes. P(N) = PW | A). 0 Yes. The two events can occur together. 0 No. P(N) z P(N IA). (f) Compute P(A or 5). (Enter your answer as a fraction.) m or s) = Z Need Help? (b) Are the events 5 = sale and Pa = passive approach independent? Explain. 0 Yes. P(S) = P(S | Pa). 0 Yes. The two events can occur together. 0 No. P(S) a: P(S | Pa). O No. The two events cannot occur together. (c) Compute PM and S) and P(Pa and 5). (Enter your answers as fractions.) Pm s> =: m 5) =: (d) Compute P(N) and P(N | A). (Enter your answers as fractions.) PM!) =: PM I A) =|:I (e) Are the events N = no sale and A aggressive approach independent? Explain. O No. The two evens cannot occur together. 0 Yes. P(N) = PW | A). 0 Yes. The two events can occur together. 0 No. P(N) z P(N IA). (f) Compute P(A or 5). (Enter your answer as a fraction.) m or s) = Z Need Help? Are customers more loyal in the East or in the West? The following table is based on information from a recent study. The columns represent length of customer loyalty (in years) at a primary supermarket. The rows represent regions of the United States. Less Than 1 Year 1-2 Years 3-4 Years 59 Years 10-14 Years 15 or More Years Row Total East 32 38 59 112 77 120 438 Midwest 31 72 68 120 63 159 513 South 53 110 93 158 106 208 728 West 41 71 67 78 45 93 395 Column Total 157 291 287 468 291 580 2074 What is the probability that a customer chosen at random has the following characteristics? (Enter your answers as fractions.) (a) has been loyal 10 to 14 years (b) has been loyal 10 to 14 years, given that he or she is from the East (c) has been loyal at least 10 years (d) has been loyal at least 10 years, given that he or she is from the West (e) is from the West, given that he or she has been loyal less than 1 year (f) is from the South, given that he or she has been loyal less than 1 year (9) has been loyal 1 or more years, given that he or she is from the East