{ "key_pair_value_system": true, "answer_rating_count": "", "question_feedback_html": { "html_star": "", "html_star_feedback": "" }, "answer_average_rating_value": "", "answer_date_js": "2024-06-28T05:30:46-04:00", "answer_date": "2024-06-28 05:30:46", "is_docs_available": null, "is_excel_available": null, "is_pdf_available": null, "count_file_available": 0, "main_page": "student_question_view", "question_id": "4244724", "url": "\/study-help\/questions\/jen-and-fiona-play-uno-together-to-see-who-can-4244724", "question_creation_date_js": "2024-06-28T05:30:46-04:00", "question_creation_date": "Jun 28, 2024 05:30 AM", "meta_title": "[Solved] Jen and Fiona play Uno together to see wh | SolutionInn", "meta_description": "Answer of - Jen and Fiona play Uno together to see who can win more. Over the past 4 years, Jen's amount of wins have been approx. | SolutionInn", "meta_keywords": "jen,fiona,play,uno,win,4,years,s,amount,wins,approx,distributed", "question_title_h1": "Jen and Fiona play Uno together to see who can win more. Over the past 4 years, Jen's amount of wins have been approx. normally", "question_title": "Jen and Fiona play Uno together to see who can win more.", "question_title_for_js_snippet": "Jen and Fiona play Uno together to see who can win more Over the past 4 years, Jen's amount of wins have been approx normally distributed with mean 494 and standard deviation 110 Fiona's are approx normal as well with mean 501 and standard deviation 112 What's the probability that Jen wins more games than Fiona if the games are all independent What's the probability that Fiona wins more games than Jen if the games are all independent What's the probability that Fiona wins TWENTY more games than Jen if the games are all independent ", "question_description": "
Jen and Fiona play Uno together to see who can win more. Over the past 4 years, Jen's amount of wins have been approx. normally distributed with mean 494 and standard deviation 110. Fiona's are approx. normal as well with mean 501 and standard deviation 112.<\/p>