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dent/PlayerTest.aspx?testid-1771909928centerwineyes https://www.n To see favorites here, select then , and drag to the Favorites Bar folder, or import from Distance Statistics Spring 2018 tl Test: EXAM 1 - CHAPTERS 1 & 2 100121 (13 complete) This Question: 1 pt Use the frequency distribution High Temperatures (F) dsfibution shown below to construc an expanded trequency Class-120-30 31-4142-52 53-6164-74T5 85 1 86-96 Frequency. fl 17 43 I 60 l- 74 68 Complete the table below nt Temperatures (E) (Round to the nearest hundredth as needed) umulative frequency ch euency C 17 31-41 43 42-5268 or Success53-6369 64-7474 68 nase Options 86-9626 0-30 Contents31-41 edia Library5474 75-85 rson Tuto Enter your answer in each of the answier boxes 0 Type here to search Quizzes &Tests; -

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Solution:-

class frequency ,f midpoint relative frequency cumulative frequency 20 - 30 17 25 17/365 = 0.05 0.05 31 - 41 43 36 43/365 = 0.12 0.16 42 - 52 68 47 68/365 = 0.18 0.34 53 - 63 69 58 69/365 = 0.19 0.53 64 - 74 74 69 74/366 = 0.20 0.73 75 - 85 68 80 68/365 = 0.19 0.92 86 - 96 24 91 24/365 = 0.07 0.99 = 1.00

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