Data Analysis
The following are the final scores in Probability and Statistics for 40 selected Year 1 Computer Science students during the academic year 2023:
30, 83, 90, 83, 75, 45, 90, 90, 68, 83, 58, 83, 73, 78, 90, 83, 53, 70, 55, 35,
31, 45, 64, 73, 65, 45, 80, 80, 68, 73, 48, 73, 73, 78, 80, 63, 43, 60, 45, 55.<\/p>\n\n\n\n
(a) Organize the data into an appropriate table and create a corresponding graph.
(b) Identify the mode of the data set.
(c) Compute the following summary measures:<\/p>\n\n\n\n
Arithmetic mean
Median
Variance
Standard deviation<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n (Imagine a bar chart with frequency on the y-axis and score ranges on the x-axis.)<\/p>\n\n\n\n The mode is the most frequently occurring score. From the dataset:<\/p>\n\n\n\n Thus, the dataset is bimodal<\/strong> with modes 73 and 83<\/strong>.<\/p>\n\n\n\n [ Arranging the data in ascending order:<\/p>\n\n\n\n 30, 31, 35, 43, 45, 45, 45, 45, 48, 53, 55, 55, 58, 60, 63, 64, 65, 68, 68, 70, 73, 73, 73, 73, 73, 73, 75, 78, 78, 80, 80, 80, 83, 83, 83, 83, 83, 90, 90, 90<\/p>\n\n\n\n For 40 numbers<\/strong>, the median is the average of the 20th and 21st values:<\/p>\n\n\n\n [ Using:<\/p>\n\n\n\n [ [ [ The analysis of the final scores of 40 students in Probability and Statistics reveals key insights about their performance. Organizing the data into a frequency distribution table and histogram helps in visualizing how scores are distributed. Most students scored between 70 and 79, with relatively fewer students scoring below 50.<\/p>\n\n\n\n The mode, which represents the most frequently occurring scores, is both 73 and 83<\/strong>, making this dataset bimodal<\/strong>. This suggests that students’ performance clustered around these two scores.<\/p>\n\n\n\n The mean (65.375)<\/strong> is a measure of central tendency, giving an idea of the average score of all students. However, the median (71.5)<\/strong> is slightly higher, indicating a right-skewed distribution, where a few lower scores pull the mean down.<\/p>\n\n\n\n To measure score variability, the variance (354.2)<\/strong> and standard deviation (18.83)<\/strong> were computed. A high standard deviation indicates that the scores are widely spread out from the mean, showing performance disparity among students.<\/p>\n\n\n\n Overall, this statistical analysis helps educators understand student performance, identify common score ranges, and assess whether interventions are needed to support students struggling in Probability and Statistics.<\/p>\n","protected":false},"excerpt":{"rendered":" Data AnalysisThe following are the final scores in Probability and Statistics for 40 selected Year 1 Computer Science students during the academic year 2023:30, 83, 90, 83, 75, 45, 90, 90, 68, 83, 58, 83, 73, 78, 90, 83, 53, 70, 55, 35,31, 45, 64, 73, 65, 45, 80, 80, 68, 73, 48, 73, 73, […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-202199","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202199","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=202199"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202199\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202199"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202199"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202199"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Frequency Distribution Table and Graph<\/h3>\n\n\n\n
Frequency Distribution Table:<\/h4>\n\n\n\n
Score Range<\/th> Frequency (f)<\/th><\/tr><\/thead> 30 – 39<\/td> 2<\/td><\/tr> 40 – 49<\/td> 5<\/td><\/tr> 50 – 59<\/td> 5<\/td><\/tr> 60 – 69<\/td> 5<\/td><\/tr> 70 – 79<\/td> 9<\/td><\/tr> 80 – 89<\/td> 7<\/td><\/tr> 90 – 99<\/td> 7<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Histogram:<\/h4>\n\n\n\n
\n\n\n\n(b) Mode<\/h3>\n\n\n\n
\n
\n\n\n\n(c) Summary Measures<\/h3>\n\n\n\n
Arithmetic Mean (\ud835\udf07)<\/strong><\/h4>\n\n\n\n
\\mu = \\frac{\\sum X}{N} = \\frac{30+83+90+\u2026+55}{40} = \\frac{2615}{40} = 65.375
]<\/p>\n\n\n\nMedian<\/strong><\/h4>\n\n\n\n
\\frac{70 + 73}{2} = 71.5
]<\/p>\n\n\n\nVariance (\u03c3\u00b2)<\/strong><\/h4>\n\n\n\n
\\sigma^2 = \\frac{\\sum (X – \\mu)^2}{N}
]<\/p>\n\n\n\n
\\sigma^2 = \\frac{(30-65.375)^2 + (83-65.375)^2 + \\dots + (55-65.375)^2}{40} = 354.2
]<\/p>\n\n\n\nStandard Deviation (\u03c3)<\/strong><\/h4>\n\n\n\n
\\sigma = \\sqrt{354.2} = 18.83
]<\/p>\n\n\n\n
\n\n\n\nExplanation (300 Words)<\/h3>\n\n\n\n