Question

The following frequency distribution shows the number of passengers who flew on a small regional aircraft company’s flight number 1853 for the past 50 days. Number of passengers on Flight 1853 Frequency (days) Relative Frequency Cumulative Relative Frequency 50 to 59 5 60 to 69 13 70 to 79 17 80 to 89 7 90 to 99 6 100 to 109 2 Use this data to construct a relative frequency and cumulative relative frequency distribution. Report all of your answers in the table above to 2 decimal places, using conventional rounding rules. What percent of the days are there less than 90 passengers on Flight 1853? Report your final answer to 2 decimal places, using conventional rounding rules. ANSWER: 84 % What proportion of the days are there 60 – 69 passengers on Flight 1853? Report your final answer to 2 decimal places, using conventional rounding rules. ANSWER: 26 What percent of the days are there no more than 79 passengers on Flight 1853? Report your final answer to 2 decimal places, using conventional rounding rules. ANSWER: 70 % What is the approximate mean number of passengers on Flight 1853? Report your final answer to 2 decimal places, using conventional rounding rules. ANSWER: What is the approximate variance of the number passengers on Flight 1853? Report your final answer to 2 decimal places, using conventional rounding rules. ANSWER: What is the midpoint of the class interval with the largest frequency? Report your final answer to 2 decimal places, using conventional rounding rules. ANSWER:

Answer 1

| Cumulative Number of Cumulative | Relative days relative воалес pa, squaps #т алхам I frequenly Frequeny 50-59 O. 10 (10%) 5 1(10%) 0.10 60-69 13 0.26 (264) 18 | 1364) 0:36 70-19 0-34 (342) | (70%) 0.70 80-89 0.1411415 42 | (847.) 0.84 90-99 6 0.12 (141 48 |(96%) 0.96 100-109 , 2 , 01 049) 50 (100%) 1000. . Relative frequency = frequening vann e Total frequency 550). » The percent of the days are there less than 90 paxsangers on flight 1853 = 84% = 84.00% » The proportion of the dam are there 60-69 passangers on Flight 1853 = 26% or 20.26). » The percentage of the days there no more than 79 paprangers on Flight 1853 = 70% 20170) (Frem curtalarkive seletive frequency) - Approximate mean number of parsangers on Flight 1853 = [74.907 meus sit 374.5 = 74.90 1 (this? (Elfy mile) -> variance number of passangers = -1 variance = H (288692.5 – (3745)') 167.jpg |167-18 | Answer - mid point of the class interval with the largest Frequency is = 70+79 = 54.50) are SO