mylab.pearson.com This test: 14 pointlsl pOSSIbie (6'3 Question 2, 6.5.169 This question: 1 Submit test pomtis) pOSSible The noshow rate for an airline's reservations is 16%. In other words, the probability is 0.16 that a person making a reservation will not take the ight. For the next ight, 44 people have reservations. Use the normal approximation tn the binomial to approximate the following probabilities. Complete pans (a) through (e). Click here to view Page 1 of the table of areas under the standard normal curve. Click here to view Page 2 of the table of areas under the standard normal curve. a. What is the probability that exactly 5 people do not take the ight? P(X = 5) = Cl (Round to four decimal places as needed.) I). What is the probability that between 8 and 12 people, inclusive, do not take the ight? P(8 s X s 12) = [3 (Round to tour decimal places as needed.) c. What is the probability that at least 1 person does not take the ight? P(X21)= D (Round to four decimal places as needed.) d. What is the probability that at most 2 people do not take the ight? P(X s 2) = D (Round to four decimal places as needed.) 9. Comment on the accuracy of the normal approximation in this case. O A. For P(X = 5) and P(B s X 512), the binomial distribution is not bell-shaped, so the normal approximation is not valid. O B. For P(Xz 1) and P(XSZ), np is less than 5, so the normal approximation is not valid. 0 C. In all cases. rip and n(1 - p) are both 5 or greater. so the normal approximation is valid. 0 D. For P(Xz 1) and P(XSZ), n(1 - p) is less than 5. so the normal approximation is not valid. Done 3 mylab.pearson.com AA 6 possible This question: 1 pointls'l possible Question 6, 3.1.27 Submit test The data below are the frequency of cremation burials found in 17 archaeological sites. ' a. Obtain the mean, median, and mode of these data. b. Which measure of center do you think works best here? 81 69 43 47 530 37 36 263 2552 E 43 361 29 86 419 53 233 138 (E) a. The mean is D. (Round to one decimal place as needed.) The median is El. (Round to one decimal place as needed.) Select the correct choice below and ll in any answer boxes within your choice. O A- The mode is . (Round to one decimal place as needed.) O B. There is no mode. b. Which measure of center works best here? Q A. Each center of measure works equally well. 0 B. The mode works best. The median is highly affected by the largest value and the mean is well to the left of the distribution's center. 0 C. The median works best. The mean is highly affected by the largest value and the mode is well to the left of the distribution's center. 0 D. The mean works best. The mode is highly affected by the largest value and the median is well to the left of the distribution's center. Done 6 mylab.pearson.com AA 6 I possible \"7 This question: 1 Submit test pomtisi possmle Question 6, 6.5.169 The noshow rate for an airtine's reservations is 16%. In other words, the probability is 0.16 that a person making a reservation will not take the ight. For the next flight. 44 people have reservations. Use the normal approximation to the binomial to approximate the following probabilities. Complete parts (a) through (e). Click here to view Page 1 of the table of areas under the standard normal curve. Click here to view Page 2 of the table of areas under the standard normal curve. a. What is the probability that exactly 4 people do not take the ight? P(X = 4) = [3 (Round to four decimal places as needed.) b. What is the probability that between 9 and 12 people, inclusive, do not take the flight? P(Q SXS 12) = [I (Round to tour decimal places as needed.) c. What is the probability that at least 1 person does not take the ight? P(X 21): D (Round to four decimal places as needed.) d. What is the probability that at most 2 people do not take the ight? P(X S 2) = D (Round to four decimal places as needed.) a. Comment on the accuracy of the normal approximation in this case. O A. For P(X 21) and P(X s 2), n(1 p) is less than 5. so the normal approximation is not valid. 0 B. For P(Xz 1 ) and P(st), np is less than 5. so the normal approximation is not valid. 0 C. For P(X = 4) and HQ 5 X512), the binomial distribution is not bell-shaped. so the normal approximation is not valid. 0 D. In all cases, np and n(1 - p) are both 5 or greater, so the normal approximation is valid. Done 6 mylab.pearson.com AA 6 possible \"7 This question: 1 Submit test Question 10' 4.2.65 potntts) possible The following table provides a frequency distribution tor a. Determine the number of outcomes (housing I6 the number of rooms in this country's housing units. The units) that constitute the event (not A). frequencies are in thousands, The number of housing units in event (not A) is Rooms No. of units at D. 1 553 2 1,414 b. Determine the number of outcomes (housing 3 10345 units) that constitute the event (A & B). 4 23,396 5 27,932 The number of housing units in event (A 8: B) is 6 24,605 [:I 7 14,679 8 T 17268 1:. Determine the number of outcomes (housing units that constitute the event C or D . A = event that the unit has at most 4 rooms ) ( ) B = event that the UN\" has at least 2 rooms The number of housing uniL'; in event (C or D) is C = event that the unit has between 5 and 7 rooms. El inclusive D = event that the unit has more than 7 rooms