[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-465310-105":59,"doc-detail-465310-en":130},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":123,"head_meta":125,"extra_data":127,"updated_unix":129},105,"en","statistics-1-normal-distribution-exam-questions","Statistics 1 - Normal Distribution Exam Questions","","This collection of examination questions and mark schemes focuses on the application of the normal distribution in statistical analysis. The document provides diverse practical problems involving continuous random variables, such as calculating probabilities for weights of red gravel bags, determining installation times for satellite dishes by different technicians, and analyzing volumes of emulsion paint. These exercises are designed to test a student's proficiency in using mean and standard deviation parameters to compute probabilities, identify threshold weights or times for specified proportions, and solve for distribution variables. Each section is structured to assess foundational understanding through to more complex statistical modeling techniques, providing essential practice for students preparing for advanced mathematics assessments.",{"@graph":69,"@context":122},[70,84,105],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/statistics-1-normal-distribution-exam-questions/465310/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/statistics-1-normal-distribution-exam-questions/465310.png","ImageObject",300,407,{"name":92,"@type":93},"Ophelia","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-30","2026-09-30",true,{"@type":102,"interactionType":103,"userInteractionCount":8},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"How do you determine the probability that a bag of red gravel weighs less than 25 kg given a mean of 25.8 kg and standard deviation of 0.5 kg?","Question",{"text":112,"@type":113},"To find this probability, you calculate the z-score using the formula z = (x - μ) / σ, which yields z = (25 - 25.8) / 0.5 = -1.6, then find the corresponding area under the standard normal distribution curve.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"What approach is used to find the time exceeded by 10% of installations for a normally distributed variable with a mean of 134 and standard deviation of 16?",{"text":117,"@type":113},"This requires finding the value of x such that P(X > x) = 0.10, which is equivalent to P(X \u003C x) = 0.90. You find the z-value corresponding to a cumulative probability of 0.90 (approximately 1.282) and then calculate x = μ + zσ = 134 + 1.282 * 16.",{"name":119,"@type":110,"acceptedAnswer":120},"If a normal distribution has a mean of 10.25 and a variance of σ², how do you find the value of σ for a specific tail probability?",{"text":121,"@type":113},"For a condition like 98% of values being above 10, you identify the z-score for the 2nd percentile (the point where 98% is above), set up the equation (10 - 10.25) / σ = z_score, and solve for σ.","https://schema.org",{"og:url":83,"og:type":124,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":126,"canonical":83},"index,follow",{"doc_id":128,"site_id":62},465310,1790769463,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":8,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":139,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":129,"read_time":144},7971461741311,"https://ap-avatar.wpscdn.com/avatar/74000253aff267980c6?x-image-process=image/resize,m_fixed,w_180,h_180&k=1779345379180704826","# Statistics 1 – Normal Distribution Exam Questions\n\nPrepared by Toot Hill School Maths Dept November 2007 1  \nPrepared by Toot Hill School Maths Dept November 2007 1  \nPrepared by Toot Hill School Maths Dept November 2007 1  \nPrepared by Toot Hill School Maths Dept November 2007 1  \n# Statistics 1 – Normal Distribution Exam Questions Mark Scheme\n\nPrepared by Toot Hill School Maths Dept November 2007 1  \nPrepared by Toot Hill School Maths Dept November 2007 1  \nPrepared by Toot Hill School Maths Dept November 2007 1  \nPrepared by Toot Hill School Maths Dept November 2007 1  \nPrepared by Toot Hill School Maths Dept November 2007 1  \nPrepared by Toot Hill School Maths Dept November 2007 1","cbCaimsrG7w1wqp7","https://ap.wps.com/l/cbCaimsrG7w1wqp7","pdf",1820752,14,"English","# Normal Distribution Exam Questions\n## Gravel Weight Modeling\n## Satellite Dish Installation Time Analysis\n## Emulsion Paint Volume Distribution","[{\"question\":\"How do you determine the probability that a bag of red gravel weighs less than 25 kg given a mean of 25.8 kg and standard deviation of 0.5 kg?\",\"answer\":\"To find this probability, you calculate the z-score using the formula z = (x - μ) / σ, which yields z = (25 - 25.8) / 0.5 = -1.6, then find the corresponding area under the standard normal distribution curve.\"},{\"question\":\"What approach is used to find the time exceeded by 10% of installations for a normally distributed variable with a mean of 134 and standard deviation of 16?\",\"answer\":\"This requires finding the value of x such that P(X \\u003e x) = 0.10, which is equivalent to P(X \\u003c x) = 0.90. You find the z-value corresponding to a cumulative probability of 0.90 (approximately 1.282) and then calculate x = μ + zσ = 134 + 1.282 * 16.\"},{\"question\":\"If a normal distribution has a mean of 10.25 and a variance of σ², how do you find the value of σ for a specific tail probability?\",\"answer\":\"For a condition like 98% of values being above 10, you identify the z-score for the 2nd percentile (the point where 98% is above), set up the equation (10 - 10.25) / σ = z_score, and solve for σ.\"}]","Statistics 1 - Normal Distribution Exam Questions | PDF",35]