{"id":20733440,"url":"https://github.com/moindalvs/assignment_2_set_1","last_synced_at":"2026-02-16T16:02:04.283Z","repository":{"id":127427938,"uuid":"465741918","full_name":"MoinDalvs/Assignment_2_Set_1","owner":"MoinDalvs","description":"Assignment 2 Set 2","archived":false,"fork":false,"pushed_at":"2022-03-03T14:28:23.000Z","size":234,"stargazers_count":2,"open_issues_count":0,"forks_count":1,"subscribers_count":3,"default_branch":"main","last_synced_at":"2025-03-22T20:46:21.132Z","etag":null,"topics":["barplot","boxplot","descriptive-statistics","inference","labels","matplotlib","outlier-detection","outliers","pie-chart","probability","scipy-stats","seaborn-plots"],"latest_commit_sha":null,"homepage":"","language":"Jupyter Notebook","has_issues":true,"has_wiki":null,"has_pages":null,"mirror_url":null,"source_name":null,"license":null,"status":null,"scm":"git","pull_requests_enabled":true,"icon_url":"https://github.com/MoinDalvs.png","metadata":{"files":{"readme":"README.md","changelog":null,"contributing":null,"funding":null,"license":null,"code_of_conduct":null,"threat_model":null,"audit":null,"citation":null,"codeowners":null,"security":null,"support":null,"governance":null,"roadmap":null,"authors":null,"dei":null,"publiccode":null,"codemeta":null}},"created_at":"2022-03-03T13:59:55.000Z","updated_at":"2023-05-01T02:37:32.000Z","dependencies_parsed_at":"2023-08-16T11:25:38.873Z","dependency_job_id":null,"html_url":"https://github.com/MoinDalvs/Assignment_2_Set_1","commit_stats":null,"previous_names":[],"tags_count":0,"template":false,"template_full_name":null,"purl":"pkg:github/MoinDalvs/Assignment_2_Set_1","repository_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/repositories/MoinDalvs%2FAssignment_2_Set_1","tags_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/repositories/MoinDalvs%2FAssignment_2_Set_1/tags","releases_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/repositories/MoinDalvs%2FAssignment_2_Set_1/releases","manifests_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/repositories/MoinDalvs%2FAssignment_2_Set_1/manifests","owner_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/owners/MoinDalvs","download_url":"https://codeload.github.com/MoinDalvs/Assignment_2_Set_1/tar.gz/refs/heads/main","sbom_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/repositories/MoinDalvs%2FAssignment_2_Set_1/sbom","host":{"name":"GitHub","url":"https://github.com","kind":"github","repositories_count":264863983,"owners_count":23675271,"icon_url":"https://github.com/github.png","version":null,"created_at":"2022-05-30T11:31:42.601Z","updated_at":"2022-07-04T15:15:14.044Z","host_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub","repositories_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/repositories","repository_names_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/repository_names","owners_url":"https://repos.ecosyste.ms/api/v1/hosts/GitHub/owners"}},"keywords":["barplot","boxplot","descriptive-statistics","inference","labels","matplotlib","outlier-detection","outliers","pie-chart","probability","scipy-stats","seaborn-plots"],"created_at":"2024-11-17T05:25:25.284Z","updated_at":"2025-10-07T13:41:06.500Z","avatar_url":"https://github.com/MoinDalvs.png","language":"Jupyter Notebook","funding_links":[],"categories":[],"sub_categories":[],"readme":"# Topics: Descriptive Statistics and Probability\n\n\n\tLook at the data given below. Plot the data, find the outliers and find out  μ,σ,σ^2\n\nName of company\tMeasure X\\\nAllied Signal\t24.23%\\\nBankers Trust\t25.53%\\\nGeneral Mills\t25.41%\\\nITT Industries\t24.14%\\\nJ.P.Morgan \u0026 Co.\t29.62%\\\nLehman Brothers\t28.25%\\\nMarriott\t25.81%\\\nMCI\t24.39%\\\nMerrill Lynch\t40.26%\\\nMicrosoft\t32.95%\\\nMorgan Stanley\t91.36%\\\nSun Microsystems\t25.99%\\\nTravelers\t39.42%\\\nUS Airways\t26.71%\\\nWarner-Lambert\t35.00%\\\n\t\n  \nThe following is the outlier in the boxplot: Morgan Stanley 91.36%\nmeasure_x.describe()\nMean = 33.271333\nStandard deviation = 16.945401\nmeasure_x.var()\nVariance = 287.1466123809524\n\n\n\n\n\n\t\n \nAnswer the following three questions based on the box-plot above.\n\n\tWhat is inter-quartile range of this dataset? (please approximate the numbers) In one line, explain what this value implies.\nAns: Approximately (First Quantile Range) Q1 = 5 (Third Quantile Range) Q3 = 12, Median (Second Quartile Range) = 7\n(Inter-Quartile Range) IQR = Q3 – Q1 = 12 – 5 = 7 \nSecond Quartile Range is the Median Value\n\n\tWhat can we say about the skewness of this dataset?\nAns:  Right-Skewed median is towards the left side it is not normal distribution\n\n\tIf it was found that the data point with the value 25 is actually 2.5, how would the new box-plot be affected?\nAns: In that case there would be no Outliers on the given dataset because of the outlier the data had positive skewness it will reduce and the data will normal distributed\n\t\n \n\nAnswer the following three questions based on the histogram above.\n\tWhere would the mode of this dataset lie?\nAns: The mode of this data set lie in between 5 to 10 and approximately between 4 to 8 .\n\n\tComment on the skewness of the dataset.\t\nAns: Right-Skewed. Mean\u003eMedian\u003eMode\n\n\tSuppose that the above histogram and the box-plot in question 2 are plotted for the same dataset. Explain how these graphs complement each other in providing information about any dataset. \nAns: They both are right-skewed and both have outliers the median can be easily visualized in box plot where as in histogram mode is more visible.\n\n\n\tAT\u0026T was running commercials in 1990 aimed at luring back customers who had switched to one of the other long-distance phone service providers. One such commercial shows a businessman trying to reach Phoenix and mistakenly getting Fiji, where a half-naked native on a beach responds incomprehensibly in Polynesian. When asked about this advertisement, AT\u0026T admitted that the portrayed incident did not actually take place but added that this was an enactment of something that “could happen.” Suppose that one in 200 long-distance telephone calls is misdirected. What is the probability that at least one in five attempted telephone calls reaches the wrong number? (Assume independence of attempts.)\n\nAns:  IF 1 in 200 long-distance telephone calls are getting misdirected.  \nprobability of call misdirecting   = 1/200\nProbability of call not Misdirecting = 1-1/200 = 199/200\nThe probability for at least one in five attempted telephone calls reaches the wrong number\nNumber of Calls = 5\nn = 5\np = 1/200\nq = 199/200\nP(x) = at least one in five attempted telephone calls reaches the wrong number\nP(x) = ⁿCₓ pˣ qⁿ⁻ˣ\nP(x) = (nCx) (p^x) (q^n-x)     # nCr = n! / r! * (n - r)!\nP(1) = (5C1) (1/200)^1 (199/200)^5-1\nP(1) = 0.0245037\n\n\tReturns on a certain business venture, to the nearest $1,000, are known to follow the following probability distribution\nx\tP(x)\n-2,000\t0.1\n-1,000\t0.1\n0\t0.2\n1000\t0.2\n2000\t0.3\n3000\t0.1\nE(X) =Sum X.*P(X) | E(X^2) =X^2*P(X)\n           -200                |          400000\n          -100                 |          100000\n             0                 |             0\n           200                 |          200000  \n           600                 |          1200000\n           300                 |          900000\nTotal:     800                 |          2800000\n\tWhat is the most likely monetary outcome of the business venture?\nAns: The most likely monetary outcome of the business venture is 2000$\nAs for 2000$ the probability is 0.3 which is maximum as compared to others\n\n\tIs the venture likely to be successful? Explain\nAns:  Yes, the probability that the venture will make more than 0 or a profit\np(x\u003e0)+p(x\u003e1000)+p(x\u003e2000)+p(x=3000) = 0.2+0.2+0.3+0.1 = 0.8 this states that there is a good 80% chances for this venture to be making a profit \n\n\tWhat is the long-term average earning of business ventures of this kind? Explain\nAns:  The long-term average is Expected value = Sum (X * P(X)) = 800$ which means on an average the returns will be + 800$\n\n\tWhat is the good measure of the risk involved in a venture of this kind? Compute this measure\nAns: The good measure of the risk involved in a venture of this kind depends on the Variability in the distribution. Higher Variance means more chances of risk \nVar (X) = E(X^2) –(E(X))^2\n             = 2800000 – 800^2\n             = 2160000\n                              \n","project_url":"https://awesome.ecosyste.ms/api/v1/projects/github.com%2Fmoindalvs%2Fassignment_2_set_1","html_url":"https://awesome.ecosyste.ms/projects/github.com%2Fmoindalvs%2Fassignment_2_set_1","lists_url":"https://awesome.ecosyste.ms/api/v1/projects/github.com%2Fmoindalvs%2Fassignment_2_set_1/lists"}